Assume that $\triangle PBD$ and $\triangle PCE$ share the same incenter $\Longrightarrow$ $\angle BPD$ and $\angle CPE$ share angle bisectors $\Longrightarrow$ $\angle BPE=\angle CPD.$ Since $\angle PBA=\angle PCA,$ then we deduce that $\triangle PEB$ and $\triangle PDC$ are directly similar $\Longrightarrow$ $\angle PEA=\angle PDA$ $\Longrightarrow$ $A,D,E,P$ are concyclic. $AP,DE,BC$ concur at the radical center $R$ of $\Omega,$ $(O)$ and $\odot(AED).$ If $U \equiv BD \cap CE,$ then $AU,AR$ are the polars of $R,U$ WRT $(O)$ $\Longrightarrow$ $AU$ is perpendicular to $OR$ at $V$ and $OU$ is perpendicular to $AR$ at $P^*.$ Whence, $RE \cdot RD=RO \cdot RV=RA \cdot RP^*$ implies that $A,D,E,P^*$ are concyclic $\Longrightarrow$ $P$ and $P^*$ are identical and the conclusion follows. The converse can be proved with similar arguments. Let $U \equiv BD \cap CE$ and $R \equiv BC \cap DE.$ If $OP$ passes through $U,$ then $P$ coincides with the orthogonal projection of $O$ onto $AR$ and since the pencils $P(B,D,U,R)$ and $P(E,C,U,R)$ are harmonic, $PU$ bisects $\angle BPD$ and $\angle CPE$ $\Longrightarrow$ $O$ becomes the midpoint of the arcs $DB$ and $EC$ of $\odot(PBD)$ and $\odot(PEC).$ Thus, the intersection $\overline{UP} \cap (O)$ is the common incenter of $\triangle PBD$ and $\triangle PCE.$

Three hours for the 'if' part, two for the 'only if'. Oops. (Then again I haven't tackled a difficult geo in a while so I guess I'm going to be a bit rusty.) First, tackle the 'if' direction. Note that the condition of concurrent incenters implies that the angle bisectors of $\angle BPD$, $\angle CPE$ must be the same line, so $\angle BPE=\angle CPD$. Additionally, $\angle PBA=\angle PCD$ since $APBC$ is cyclic, so $\triangle PEB\sim\triangle PDC$. Looking at the spiral similarity sending the former to the latter gives $\triangle PED\sim\triangle PBC$, and it is well known that this implies $P=(AED)\cap\Omega$. Now let $M$ and $N$ be the midpoints of $\overline{BE}$ and $\overline{CD}$ respectively. By MGT (or spiral similarity if you prefer) $\triangle PED\sim\triangle PMN$ so $\triangle PEM\sim\triangle PDN\implies \angle PMA=\angle PNA\implies APMN$ is cyclic. But note that since $\angle OMA=\angle ONA=90^\circ$ we have $AMON$ cyclic as well. Hence $APMO$ is too cyclic (heh APMO), so $\angle APO=\angle AMO=90^\circ$. Denote $X=BD\cap CE$ and $T=DE\cap BC$. Brokard gives $O$ as the orthocenter of $\triangle AXT\implies OX\perp AT$. But since $OP\perp AT$ we must have $O,P,X$ collinear $\implies BD,CE,OP$ concurrent. Similar reasoning can be applied in the opposite direction. Let $P'$ denote the point that $AT$ intersects $\Omega$ (where $T=DE\cap BC$ is defined as before). Since $DE$ is the radical axis of $(AED)$ and $\omega$ and $BC$ is the radical axis of $\omega$ and $\Omega$, $T$ is the radical center of all three circles. Hence $AT\equiv AP'$ is the radical axis of $\Omega$ and $(AED)\implies APED$ is cyclic. Using the same reasoning as before gives $AP'\perp P'O$ and $AT\perp XO\equiv P'O$. Since two nonparallel lines can only intersect in one point, $P'$ is the point at which $OX$ intersects $\Omega\implies P'\equiv P$. Thus $\triangle PED\sim\triangle PBC$, so spiral similarity gives $\triangle PEB\sim\triangle PDC$. Now I claim that this implies the incenters are identical. To prove this, let $I$ and $I'$ be the incenters of $\triangle PDB$ and $\triangle PEC$ respectively. Note that since \begin{align*}\pi-\angle BPD&=\angle PDB+\angle PBD\\&=(\angle PDE+\angle EDB)+(\angle PBE+\angle EBD)\\&=(\angle PCB+\angle ECB)+(\angle PCD+\angle ECD)\\&=2\angle C=\angle BOD,\end{align*} we have that $PDOB$ is cyclic. Hence the bisector of $\angle BPD$ passes through $O$ and by Fact 5 we have $IO=ID=IB$. Similarly, we have $PEOC$ cyclic so $I'O=I'E=I'C$. But since it's obvious that $I$, $I'$, and $O$ are collinear, it must be true that $I\equiv I'$. $\blacksquare$

If $ BD, CE, OP $ are concurrent : Let $ OP $ cuts arc $ DE $ of $ \omega $ at $ I $ and let $ F $ $ \equiv $ $ BD $ $ \cap $ $ CE,$ $ X $ $ \equiv $ $ BC $ $ \cap $ $ DE $. Since $ AX $ is the polar of $ F $ WRT $ \omega $, so $ OF $ $ \perp $ $ AX $ $ \Longrightarrow $ $ P $ $ \equiv $ $ OF $ $ \cap $ $ \Omega $ is the Miquel point of complete quadrilateral $ \{ DE, EB, BC, CD \} $. Since $ P $ is the projection of $ O $ on $ AX $, so from $ P(B,D;F,X) $ $ = $ $ -1, $ $ P(E,C;F,X) $ $ = $ $ -1 $ we get $ PO $ bisect $ \angle BPD $ and $ \angle EPC $ . Since $ \angle OEC $ $ = $ $ \tfrac{1}{2}(180^{\circ}-\angle COE) $ $ = $ $ 90^{\circ} $ $ - $ $ \angle CBE $ $ = $ $ 90^{\circ} $ $ - $ $ \angle ADE $ $ = $ $ 90^{\circ} $ $ - $ $ \angle APE $ $ = $ $ \angle EPO $ $ = $ $\angle OPC $, so we get $ O $ is the midpoint of arc $ CE $ in $ (PCE) $, hence $ I $ is the incenter of $ \triangle PCE $. Similarly, we can get $ I $ is the incenter of $ \triangle PBD $, so $ \triangle PBD $ and $ \triangle PCE $ have the same incenter $ I $ . If $ \triangle PBD $ and $ \triangle PCE $ have common incenter : Let $ I $ be the common incenter of $ \triangle PBD, $ $ \triangle PCE $ and let $ F $ $ \equiv $ $ BD $ $ \cap $ $ CE, $ $ X $ $ \equiv $ $ BC $ $ \cap $ $ DE $. Since $ \angle BPI $ $ = $ $ \angle IPD $ and $ \angle EPI $ $ = $ $ \angle IPC $, so $ \angle BPE $ $ = $ $ \angle CPD $. Since $ \angle PBE=\angle PCD $, so $ \triangle PBE $ $ \sim $ $ \triangle PCD $ $ \Longrightarrow $ $ P $ is the Miquel point of complete quadrilateral $ \{ DE, EB, BC, CD \} $, hence we get $ P $ is the projection of $ O $ on $ AX $ and $ OP \perp AX $ ... (1). Since $ F $ is the pole of $ AX $ WRT $ \omega $, so we get $ OF $ $ \perp $ $ AX $ ... (2). From (1) and (2) we get $ O, F, P $ are collinear. i.e. $ BD, $ $ CE, $ $ OP $ are concurrent Q.E.D

Restatement: "ABCD is a cyclic quad, AB and CD cut at P, BC and AD cut at Q, BD and AC cut at R, O is center of ABCD. M is a point such that MPAD is cyclic. Prove ORM collinear iff MBD, MAC have the same incenter". Proof: if they have the same incenter then <BMC=<AMD and thus M is the miquel point, and by a well known fact, R is the inverse of M, done. If ORM are collinear then M is the miquel point, and therefore PMBC, PMAD, MBAQ, MCQD are collinear. This implies MO bisects <BMD and <AMC. Notice BRD are collinear and thus by inversion OBMD are concyclic, and thus the incenter of MBD is $ \omega \cap OR$. Same for MAC, so done.

If we see that P is on the circumference of the $ \triangle{OBD} $ and $ \triangle{OCE} $ then, the rest is easy.

We will first prove the if statement. If $PBD$ and $PCE$ have the same incenter, then $\angle BPE = \angle CPD$. Also, since $\angle PBA = \angle PAC$, it follows that there is a spiral similarity at $P$ that sends $BE$ to $CD$. Hence, $P$ is the Miquel Point of $BCDE$ and it is a well known fact that the line through the center of the $\odot (BCDE)$ and the Miquel Point passes through $BD\cap EC = K$. Now the only if statement. $P$ is the Miquel Point of $BCDE$ since it is the unique point that lies on the circumcircle of $ABC$ and line $OK$. Therefore, $PBOD$ and $PEOC$ are cyclic, and by Fact 5, the incenters of $PBE$ and $PCE$ are equal to $\omega \cap PO$. $\Box$

Suppose that $OP,BD,CE$ are concurrent. Then $P$ is the Miquel point of a cyclic quadrilateral $BEDC \implies BDOP$ is cyclic. Since $OB=OD$, by the In/Ex-Center Lemma the incenter of $\triangle PBD$ is the intersection of $(BEDC)$ with the ray $OP$. We get a similar result for $\triangle PCE$ and we're done. Now suppose that triangles $\triangle PBD$ and $\triangle PCE$ have the same incenter. Let this common incenter be $I$. Then $\angle BPE=\angle CPD=\frac{1}{2}(\angle BPD-\angle EPC)$ and $\angle PBE=\angle PBA=\angle PCA=\angle PCD$. Thus $\triangle PBE \sim \triangle PCD \implies P$ is the center of the spirals similarity taking $\overline{BE} \to \overline{CD}$, so $P$ is the center of the spiral similarity taking $\overline{BC} \to \overline{ED}$, which is the Miquel point of $BEDC \implies BD,CE,OP$ are concurrent, as desired.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10.36522563978495cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(9); defaultpen(dps); /* default pen style */ real xmin = -2.0689156638847113, xmax = 1.3861595493769383, ymin = -1.3208941057372592, ymax = 1.1618146790339252; /* image dimensions */ pen qqqqcc = rgb(0.,0.,0.8); pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pair A = (-0.2836894390492209,0.9589162122792264), B = (-0.8149553245687551,-0.5795237863600036), C = (0.7693998536964243,-0.63876745779033), D = (0.13858397801064062,0.3182684083479753), O = (-0.017615909620593166,-0.4711027685051842), P = (-0.730446411688668,0.6829700137276519), F = (-0.4404769547037505,0.21350938176220052), Q = (-0.2684630085172446,-0.0649812452274847), G = (-0.7933916669323753,-0.2573804390450106), H = (-0.2829141020605786,0.28858277751410705); draw(A--B--C--cycle, linewidth(1.2) + zzttqq); draw(arc(P,0.10024396170004786,-41.38807091410654,-22.766085970821553)--(-0.730446411688668,0.6829700137276519)--cycle, linewidth(2.) + qqwuqq); draw(arc(P,0.10024396170004786,-93.82955666991205,-75.20757172662705)--(-0.730446411688668,0.6829700137276519)--cycle, linewidth(2.) + qqwuqq); /* draw figures */ draw(circle((0.,0.), 1.), linewidth(1.2)); draw(A--B, linewidth(1.2) + zzttqq); draw(B--C, linewidth(1.2) + zzttqq); draw(C--A, linewidth(1.2) + zzttqq); draw(circle(O, 0.8046771152099158), linewidth(1.2)); draw(P--(-0.5781461904028535,0.10622821461343612), linewidth(1.2)); draw(P--D, linewidth(1.2)); draw(circle((-0.30130534866981407,0.4878134437740421), 0.47143200863434187), linewidth(1.2)); draw(P--B, linewidth(1.2)); draw(P--C, linewidth(1.2)); draw(D--B, linewidth(1.2)); draw(C--(-0.5781461904028535,0.10622821461343612), linewidth(1.2)); draw(O--P, linewidth(1.2)); draw(D--(-0.5781461904028535,0.10622821461343612), linewidth(1.2)); draw(D--F, linewidth(1.2)); draw(F--B, linewidth(1.2)); draw((-0.5781461904028535,0.10622821461343612)--F, linewidth(1.2)); draw(C--F, linewidth(1.2)); draw(P--A, linewidth(1.2)); draw(D--G, linewidth(1.2)); draw(F--H, linewidth(1.2)); draw(C--G, linewidth(1.2)); draw(O--D, linewidth(1.2)); draw((0.055020856444464084,-0.07533613256158132)--(0.06594721194558341,-0.07749822759562781), linewidth(1.2)); draw(O--G, linewidth(1.2)); draw((-0.4069829442291494,-0.3696106886632296)--(-0.4040246323238191,-0.35887251888696536), linewidth(1.2)); draw(O--H, linewidth(1.2)); draw((-0.15552273160272914,-0.09309610406486817)--(-0.1450072800784423,-0.0894238869262089), linewidth(1.2)); draw(O--(-0.5781461904028535,0.10622821461343612), linewidth(1.2)); draw((-0.30187671374276553,-0.18631666429656585)--(-0.29388538628068084,-0.17855788959518254), linewidth(1.2)); /* dots and labels */ dot(A,qqqqcc); label("$A$", (-0.2712072840638529,0.9913999441438439), NE * labelscalefactor,qqqqcc); dot(B,qqqqcc); label("$B$", (-0.9027442427741544,-0.6091619776669196), NE * labelscalefactor,qqqqcc); dot(C,qqqqcc); label("$C$", (0.8147689676866656,-0.6726498200769498), NE * labelscalefactor,qqqqcc); dot((-0.5781461904028535,0.10622821461343612),qqqqcc); label("$E$", (-0.6822075270340491,0.09588721962341673), NE * labelscalefactor,qqqqcc); dot(D,linewidth(4.pt) + qqqqcc); label("$D$", (0.17989054358636247,0.34649712387353626), NE * labelscalefactor,qqqqcc); dot(O,linewidth(4.pt) + qqqqcc); label("$O$", (0.0027928779162779143,-0.5356497390868845), NE * labelscalefactor,qqqqcc); dot(P,linewidth(4.pt) + qqqqcc); label("$P$", (-0.7857929541240986,0.7006924552137053), NE * labelscalefactor,qqqqcc); dot(F,linewidth(4.pt) + qqqqcc); label("$F$", (-0.42825615739392786,0.23957023139348527), NE * labelscalefactor,qqqqcc); dot(Q,linewidth(4.pt) + qqqqcc); label("$Q$", (-0.3547439188138928,-0.08455191143666935), NE * labelscalefactor,qqqqcc); dot(G,linewidth(4.pt) + qqqqcc); label("$G$", (-0.8492807965341289,-0.23825931937674266), NE * labelscalefactor,qqqqcc); dot(H,linewidth(4.pt) + qqqqcc); label("$H$", (-0.2712072840638529,0.3164239353635219), NE * labelscalefactor,qqqqcc); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Nice Geo the if part is pretty obvious because $\angle DPC=\angle EPB$ then we also have that $\angle PCA=\angle ABP$. So $P$ is the miquel point of $BCDE$ then $DB,CE,PO$ are concurrent. the only if part proof is exactly the backward proof of the first part. name the intersection of $BD$ and $CE$ as $Q$, Then $P$ is the intersection of $OQ$ with $(ABC)$ then again we observe that $P$ is the miquel point of $DEBC$. it is trivial that $\angle EPC$ and $\angle BPC$ share the same angle bisector. let the angle bisector of $\angle EPC$ intersect the circle $\Omega$ at $F$.the remaining is to show that $CF$ bisects $\angle PCE$. $G$ is the second intersection of line $PB$ with $\omega$. it is easy to see that $\triangle ODP=\triangle OGP$ also $\triangle OHP=\triangle OEP$, So $FH=FE$ and similarly $FG=FD$ which finishes the proof.

Nice Miquel point review. [asy][asy] size(7cm); defaultpen(fontsize(10pt)+linewidth(0.4)); dotfactor *= 1.5; pair A = dir(105), B = dir(210), C = dir(330), O = (B+C)/2+dir(270)*0.1, D = 2*foot(O,A,C)-C, E = 2*foot(O,A,B)-B, Q = extension(B,D,C,E), P = intersectionpoint(unitcircle,Q--Q+dir(O--Q)*69), K = intersectionpoint(P--Q,circumcircle(C,B,D)), K1 = 2O-K; draw(A--B--C--A^^B--D^^C--E, heavyblue); draw(circumcircle(A,E,D)^^unitcircle, blue+cyan); draw(circumcircle(B,C,D), purple); draw(K1--P, heavyred); draw(C--P--E^^B--P--D, heavygreen); draw(circumcircle(O,B,D)^^circumcircle(O,C,E), linewidth(0.8)+heavygreen+dotted); dot("$A$", A, dir(90)); dot("$B$", B, dir(190)); dot("$C$", C, dir(350)); dot("$O$", O, dir(265)); dot("$D$", D, dir(30)); dot("$E$", E, dir(180)); dot("$Q$", Q, dir(85)); dot("$P$", P, dir(135)); dot("$K$", K, dir(85)); dot("$K'$", K1, dir(310)); [/asy][/asy] If $\triangle PBD$ and $\triangle PCE$ share the same incenter, then we must have $$\angle BPE = \angle CPD \iff \angle BPE + \angle PBA = \angle CPD + \angle PCA \iff \angle PEA = \angle PDA,$$which forces $P \ne A = (ADE) \cap (ABC)$ to be the only candidate, so it remains to show that this $P$ works. Let $Q = \overline{BD} \cap \overline{CE}$ and suppose that line $\overline{PQ}$ meets $\omega$ at points $K$, $K'$ with $PK < PK'$. By definition $P$ is the Miquel point of cyclic quad $BCDE$, so recall the well-known properties $P$ is the inverse of $Q$ with respect to $\omega$, and $PBOD$ and $PCOE$ are cyclic. The former implies $O \in \overline{PQ}$, and the latter implies $\overline{PQ}$ bisects both $\angle BPD$ and $\angle CPE$ because $O$ is the arc midpoint. It follows by Fact 5 that $K$ is the common incenter of $\triangle PBD$ and $\triangle PCE$ (and $K'$ is a common excenter). $\blacksquare$

By Miquel point properties, $P$ is the Miquel point of $CDEB$ iff lines $OP,BD,CE$ concur, $P$ is on $(ABC)$, and $P,A$ are on the same side of $\triangle ABC$. Thus it suffices to show $P$ is the Miquel point of $CDEB$ iff the incenters of $\triangle PBD$ and $\triangle PCE$ are the same. For the only if part, let $PO$ intersect $(CDEB)$ at $I$ between $P$ and $O$. Let the antipode of $I$ wrt $(CDEB)$ be $I'$ and $PO\cap BD\cap CE=K$. As $(PK;II')$ is harmonic by Miquel point results and $\angle I'BI=\angle I'EI=\angle I'DI=\angle I'CI$, then $I$ is the incenter of both $\triangle PCE$ and $\triangle PBD$ as desired. For the if part, let $I$ be the common incenter. Observe first that $P$ lies on $(ABC)$. Then, note that \[\measuredangle EAD=\measuredangle BAC=\measuredangle BPC=\measuredangle BPI+\measuredangle IPC=\measuredangle EPI+\measuredangle IPD=\measuredangle EPD,\]which implies $A$ is the Miquel point, as desired.

Note that the result is equivalent to showing that $P$ is the miquel point of $BEDC$ if and only if $\triangle PBD$ and $\triangle PCE$ share an incenter. It is now well-known that $P$ lies in $(ODB)$ and $(OEC)$. Let line $OP$ meet $\omega$ at a point $I$; $I$ is the incenter by incenter-excenter lemma, as $O$ is the arc midpoint of $(PDB)$ and $(PEC)$. Conversely, suppose $\triangle PDB$ and $\triangle PEC$ share an incenter $I$. Now $$\angle DPE = \angle DPI + \angle EPI = \angle BPI + \angle CPI = \angle BPC = \angle BAC = \angle EAD,$$so $P$ lies on $(ADE)$. Thus $P$ is the miquel point of $BEDC$, as desired.

Let $Q = BD \cap CE$. First, we assume $O, P, Q$ are collinear. Because $BCDE$ is cyclic, Master Miquel yields $P$ as the Miquel Point of $BCDE$. Thus, we know $P$ and $Q$ are inverses wrt $\omega$. It follows that $PBOD$ and $PCOE$ are cyclic. Since $O$ is the midpoint of minor arcs $BD$ and $CE$, the Incenter-Excenter Lemma implies the shared incenter of $PBD$ and $PCE$ is the intersection between segment $OP$ and $\omega$. Now, we assume $PBD$ and $PCE$ have the same incenter $I$. This gives $$\measuredangle BPE = \measuredangle BPI - \measuredangle EPI = \measuredangle IPD - \measuredangle IPC = \measuredangle CPD.$$In addition, we have $$\measuredangle PBE = \measuredangle PBA = \measuredangle PCA = \measuredangle PCD$$so $PBE \overset{+}{\sim} PCD$. Hence, $P$ is indeed the Miquel Point of $BCDE$, which finishes via Master Miquel. $\blacksquare$ Something Random I Noticed While Solving This Problem: Let $ABC$ be a triangle and $X$ be a point on the internal or external bisector of $\angle BAC$. Then, we have $\measuredangle BAX = \measuredangle XAC$.

Let $F$ be the intersection of $BD$ and $CE$. If the three lines concur, then $P$ lies on both $(ABC)$ and $OF$, so it is the Miquel point of $BCDE$. Consequently, $PBOD$ and $PEOC$ are cyclic. From $PBOD$ cyclic, since $O$ is the arc midpoint of $BD$ in $(PBOD)$, the intersection of $PO$ and $\omega$ closer to $P$ is the incenter of $\triangle PBD$. However, we can also apply the same logic to $\triangle PEC$, giving the same incenter, which solves this part. If $\triangle PBD$ and $\triangle PEC$ share the same incenter $I$, then $\angle BPE=\angle DPC$ since $PE$ and $PC$ are isogonal in $\triangle PBD$. Furthermore, $\angle PBE=\angle PCD$ since $PACB$ is cyclic, so $\triangle PBE$ is spirally similar to $\triangle PCD$, and hence $P$ is the Miquel point of $EDCB$ which finishes.

For the "if" part, We have $\triangle PBD$ and $\triangle PCD$ with the same incentre.Call it $I$.Let $ED \cap BC=F$.Now see that $\{PB,PD\},\{PC,PE\}$ reflections around $PI$.Applying DDIT on $BDCE$ we have $\{PB,PD\},\{PC,PE\},\{PA,PF\}$ swapped.So $PA,PF$ are isogonal wrt $\angle CPE$.So we have, $$\angle FBE=180-\angle ABC=180-\angle APC=180-\angle FPE$$Which gives $(FPEB)$ cyclic.So $P$ is the miquel point of $BEDC$ by well known properties we have that $OP,BD,CE$ concurrent $\blacksquare$ For the only if part, By definition $P$ is the miquel point of $BECD$.It is well known that $PBOD$ and $PCOE$ cyclic.By fact 5 we have the incentre of $PBD$ and $PCE$ as $OP \cap \omega$ since $OB=OD=OE=OC$.Thus we have a common incentre completing the proof $\blacksquare$