Note that the locus is the inverse of \(\omega\) with center as \(X\) and radius is \(XB= XA\).

@above, I copied this verbatim from the original question paper

Cute. MathPassionForever wrote: Let $ABCD$ be a cyclic quadrilateral such that $AD=BD=AC$. A point $P$ moves along the circumcircle $\omega$ of triangle $ABCD$. The lined $AP$ and $DP$ meet the lines $CD$ and $AB$ at points $E$ and $F$ respectively. The lines $BE$ and $CF$ meet point $Q$. Find the locus of $Q$. Let $X$ be the intersection of $AC$ and $BD$. Claim 1. $BX$ is tangent to $\odot(BPF)$ and $CX$ is tangent to $\odot(CPE)$. (Proof): $\angle XBF = \angle XBA = \angle BAD = \angle BPD$ and we get $BX$ is tangent to $\odot(BPF)$ similarly $CX$ is tangent to $\odot(CPE)$. $~\square$ Let $Q'$ be the second intersection of $\odot(BPF)$ and $\odot(CPE)$. Claim 2. $C,Q',F$ and $B,Q',E$ are collinear or $Q'\equiv Q$. (Proof): From claim 1, $PX$ is the radical axis of the two circles, so $Q'$ lies on $PX$. We have \[\angle FQ'P = \angle FBP = \angle ACP = \angle XCP = \angle CEP = \angle CQ'P\]so $C,Q',F$ are collinear. Similarly $B,Q',E$ are collinear which gives us $Q'\equiv Q$. $~\square$ Now we have from the above claim, $\odot(BQPF)$ is tangent to $BX$, so $XB^2 = XQ\cdot XP$. Therefore $Q$ is the image of $P$ after inverting about $X$ with radius $XB^2$. As $P$ varies on a circle, $Q$ also varies on a circle which is the image obtained after inverting $\omega$ about $X$ with radius $XB^2$ and we are done. $~\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(12.cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -7., xmax = 17., ymin = -12., ymax = 7.; /* image dimensions */ pair A = (-2.,0.), D = (7.,0.), B = (0.15791697546462924,5.8468709482393155), C = (4.842083024535371,5.8468709482393155), P = (3.3855559604504535,-3.0534229262660375), X = (2.5,3.8454545454545443), F = (-6.077328644064,-11.047512330852673), Q = (2.6737298926368127,2.492020887950979); /* draw figures */ draw(A--D, blue); draw(B--C); draw(A--C, blue); draw(B--D, blue); draw(circle((2.5,1.660824477837162), 4.796700735524688), red); draw(B--P, linewidth(1.)); draw(P--C, linewidth(1.)); draw(A--B, linewidth(1.)); draw(P--D, linewidth(1.)); draw(P--X, linewidth(1.)); draw(B--Q, linewidth(1.)); draw(F--C, linewidth(1.)); draw(Q--(9.381614835911765,-6.452979772769065), linewidth(1.)); draw(circle((-6.077328644063998,-1.4496930746133374), 9.597819256239337), dashed); draw(circle((9.381614835911767,0.534652871096723), 6.98763264386579),dashed); draw(A--(9.381614835911765,-6.452979772769065), linewidth(1.)); draw((9.381614835911765,-6.452979772769065)--C, linewidth(1.)); draw(A--F, linewidth(1.)); draw(F--P, linewidth(1.)); /* dots and labels */ dot(A,dotstyle); label("$A$", (-2.8,-0.24580435905620365), NE * labelscalefactor); dot(D,dotstyle); label("$D$", (7.1,-0.27460146747697733), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (0.1,6), NE * labelscalefactor); dot(C,linewidth(4.pt) + dotstyle); label("$C$", (4.8,6.089559493514007), NE * labelscalefactor); dot(P,dotstyle); label("$P$", (3.25,-3.9), NE * labelscalefactor); dot((9.381614835911765,-6.452979772769065),linewidth(4.pt) + dotstyle); label("$E$", (9.505835138897918,-6.23560291057713), NE * labelscalefactor); dot(X,linewidth(4.pt) + dotstyle); label("$X$", (2.5,4.073761904059849), NE * labelscalefactor); dot(F,linewidth(4.pt) + dotstyle); label("$F$", (-6.5,-10.785546041059373), NE * labelscalefactor); dot(Q,linewidth(4.pt) + dotstyle); label("$Q$", (2.824905985278417,2.0003700977641437), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Here is a solution using similar triangles and angle chasing. [asy][asy] pair A = (-0.96, -0.28), D = (0.96, -0.28), O = origin, B = reflect(D, O) * A, C = reflect(A, O) * D, P = dir(280), E = extension(A, P, C, D), F = extension(D, P, A, B), Q = extension(B, E, C, F); draw(A--F^^D--E, gray(0.6)); draw(B--E^^C--F, gray(0.6)); draw(A--E^^D--F, gray(0.6)); draw(unitcircle); draw(circumcircle(B, Q, C), dotted); draw(A--B^^C--D); draw(C--A--D--B, blue); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$P$", P, dir(P)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$Q$", Q, dir(0)); [/asy][/asy] By angle chasing, it is easy to obtain $\triangle DAE \sim \triangle AFD$, so $AD^2 = DE \cdot AF$. Then $\measuredangle CAF = \measuredangle EDB$ and $\tfrac{AC}{AF} = \tfrac{DE}{DB}$, so $\triangle ACF \sim \triangle DEB$. Consequently, \[\measuredangle CQB = \measuredangle (\overline{FC}, \overline{BE}) = \measuredangle (\overline{FA}, \overline{BD}) = \measuredangle FBD = \measuredangle CBA,\]so $(BQC)$ is tangent to $\overline{AB}$. Analogously $(BQC)$ is tangent to $\overline{CD}$, so the locus is the circle tangent to $\overline{AB}$ at $B$ and tangent to $\overline{CD}$ at $C$. (It's easy to see that all these points do belong to the locus by reversing the construction.)

Moving Points MathPassionForever wrote: Let $ABCD$ be a cyclic quadrilateral such that $AD=BD=AC$. A point $P$ moves along the circumcircle $\omega$ of $ABCD$. The lined $AP$ and $DP$ meet the lines $CD$ and $AB$ at points $E$ and $F$ respectively. The lines $BE$ and $CF$ meet point $Q$. Find the locus of $Q$. Animate $P$ on $\omega$. Note that $AD \parallel BC$. Let $\gamma$ be the circle tangent to $AB$ at $B$ and $CD$ at $C$. Then $$P \mapsto E \mapsto BE \cap \gamma \text{ and } P\mapsto F \mapsto CF \cap \gamma$$are projective maps. We claim that these two points coincide, which will show that locus of $Q$ is a circle. We just need to check three positions of $P$. The result is obvious when $P=B, C$. For the third position take $P$ to be the midpoint of minor arc $AD$. As $AD=AC$, so $BA$ and $DP$ are external angle bisectors of $\angle CBD$ and $\angle BDC$. This means that $E$ is the $C$-excenter of $\triangle BCD$, or equivalently that $CE$ bisects $\angle BCD$. Similarly, $BF$ bisects $\angle CBD$. Thus, $$\angle DCE=\angle BCE=\angle CBF=\angle ABF$$which directly gives that these two lines meet on $\gamma$, as desired. EDIT: 500 posts on HSO

A different Moving Points solution: MathPassionForever wrote: Let $ABCD$ be a cyclic quadrilateral such that $AD=BD=AC$. A point $P$ moves along the circumcircle $\omega$ of triangle $ABCD$. The lined $AP$ and $DP$ meet the lines $CD$ and $AB$ at points $E$ and $F$ respectively. The lines $BE$ and $CF$ meet point $Q$. Find the locus of $Q$. Let $T = \overline{AC} \cap \overline{BD}$ and $\Phi$ be the inversion at $T$ with radius $TB=TC$. Let $\omega^*,P^*$ be the image of $\omega,P$ under $\Phi$, respectively. We claim that $\omega^*$ is the locus of $Q$, and for that, it suffices to show $Q = P^*$. It suffices to show points $B,P^*,E$ are collinear since that would similarly imply points $C,P^*,F$ are collinear and we would be done. Fix $A,B,C,D,T$ and animate $P$ on $\omega$. It is not hard to see that maps $P \mapsto P^*$ and $P \mapsto E$ are projective so consequently the map $E \mapsto P^*$ is projective. Since $B \in \omega^*$ so we just want to show this map is projection through $B$ and hence it suffices to prove this for three distinct positions of $P$. $P=B,C$. These cases are just trivial. $P=A,D$. These cases are not trivial, but are not hard either (they are just an easy angle chase). So we in fact proven this for four distinct positions of $P$ ! This completes the proof of the problem. $\blacksquare$ Remark: $\omega^*$ is also the image of $\omega$ under a homothety at $T$ with scale $\frac{-TB}{TD}$.

We prove the locus of $Q$ is the circle passing through $B$ and $C,$ with $\overline{AB}$ and $\overline{CD}$ tangent to the circle. Call this circle $\omega.$ Note that the length condition translates to $ABCD$ an isosceles trapezoid with $AD=AC=BD.$ Fix $ABCD$ and animate $P$ on $(ABCD).$ Observe the maps $P \xmapsto{D} F$ and $P \xmapsto{A} E$ are projective, so $\deg(E)=\deg(F)=1.$ It follows $\deg(Q)=1+1=2,$ Also note the map $E \xmapsto{B} Q':= \overline{BE} \cap \omega$ is projective and $Q' \mapsto \overline{CQ'} \cap \overline{AB}$ is projective again, so it suffices to verify these map concur for three positions of $P.$ But this is very easy to do, by checking $P=A,B,C,D.$

The locus is the circle tangent to $\overline{AB}$ and $\overline{AC}$ at $B,C$ respectively. Call it $\alpha.$ Let $Q_1=\overline{BE} \cap \alpha, Q_2=\overline{CF} \cap \alpha.$ As $$P \mapsto_A E \mapsto_B Q_1, P \mapsto_B F \mapsto_A Q_2,$$we simply need to show $Q_1=Q_2$ for $3$ cases by tethered moving points. The first two cases are $P=B,C,$ which are trivial. The last case is when $P$ is the arc midpoint of $\overarc{BC}.$ Note that this would mean proving that $\overline{BE}$ is an angle bisector of $\angle CBF,$ so by angle chasing this would mean showing that $\overline{BE} \perp \overline{DF},$ which would mean showing that $DE=DB,$ which is $=AD,$ however $ADE$ is isosceles because $\angle DAE=\angle AED=90-\frac{\angle CDA}{2}.\blacksquare$