[asy][asy] size(250); pair X=(4.8,-0.13),M=(4.84,3.35),N=(4.76,-3.61),B=(2.39433995702965,3.378111034976671),C=(3.7154324145870223,-3.5979934760297363),K=2*foot(M,B,X)-M,L=2*foot(N,C,X)-N,D=extension(B,K,C,N),A=extension(C,L,B,M),P=extension(A,C,B,D),Q=extension(P,foot(P,M,N),A,D),R=extension(P,Q,B,C),S=foot(X,B,C); path d=incircle(A,B,D); draw(circle(X,abs(X-M)),orange+dotted); draw(d,cyan); draw(incircle(A,D,C),cyan); draw(Q--P,red); draw(M--A--C^^N--D--B,orange); draw(A--D^^B--C,brown); draw(X--S,brown+dashed); draw(R--P--S,green); dot("$X$",X); dot("$B$",B,dir(90)); dot("$C$",C,dir(-90)); dot("$A$",A,dir(A)); dot("$D$",D,dir(D)); dot("$P$",P,dir(90)); dot("$Q$",Q,dir(Q)); dot("$R$",R,dir(R)); dot("$S$",S,dir(60)); [/asy][/asy] $\boxed{\textbf{Part 1:} \text{ A circle } \Gamma \text{ centered at } X \text{ is tangent to segments } AC,BD \text{ and extension of rays } AB,CD.}$ Let $s(UVW)$ denote semi-perimeter of $\triangle UVW$. Claim 1: $s(DAB) = AB+CD$ and similarly $s(ADC) = CD + AB$. Proof: As $ABPQ$ is circumscribed, so incircle of $\triangle DAB$ is tangent to $PQ$. As $PQ \parallel AB$, so the homothety at $D$ taking $QP \to AB$ takes incircle to $D$-excircle in $\triangle DAB$. It follows \begin{align*} \frac{DP}{DB} = \frac{\text{inradius } \triangle DAB}{\text{D-exradius } \triangle DAB} &=\frac{\text{area } \triangle DAB}{\text{D-exradius } \triangle DAB} \div \frac{\text{area } \triangle DAB}{\text{inradius } \triangle DAB} = \frac{s(DAB) - AB}{s(DAB)} \\ \implies \frac{AB}{s(DAB)} &= \frac{BP}{BD} = \frac{AB}{AB + CD} \end{align*}and our Claim follows. $\square$ Corollary 2: $AB + DB = AC + DC$. Proof: From Claim 1 we get \begin{align*} s(DAB) = s(ADC) &\implies AD + (AB +DB) = AD + (AC + DC) \\ &\implies AB + DB = AC + DC \end{align*}as desired. $\square$ It follows (by a version of Pitot) that there exists a circle $\Gamma$ tangent to segments $AC,BD$ and extension of rays $AB,CD$. As $X$ is the intersection of external angle bisectors of $\angle DBA, \angle DCA$, so $X$ must be the center of $\Gamma$. $\boxed{\textbf{Part 2:} \text{ (Leap of faith) Existence of } \Gamma \text{ is enough to imply } PR=PS.}$ Let $AB,DC,BD,CA$ be tangent to $\Gamma$ at $M,N,K,L$, respectively. [asy][asy] size(250); pair X=(4.8,-0.13),M=(4.84,3.35),N=(4.76,-3.61),B=(2.39433995702965,3.378111034976671),C=(3.7154324145870223,-3.5979934760297363),K=2*foot(M,B,X)-M,L=2*foot(N,C,X)-N,D=extension(B,K,C,N),A=extension(C,L,B,M),P=extension(A,C,B,D),Q=extension(P,foot(P,M,N),A,D),R=extension(P,Q,B,C),S=foot(X,B,C),T=extension(M,N,K,L); draw(circle(X,abs(X-M)),orange); draw(circle(P,abs(P-K)),cyan); draw(circumcircle(M,R,N)); draw(M--L^^N--K,brown); draw(K--T--M^^T--B,purple); draw(M--B--P--C--N,red); draw(X--S,brown+dashed); draw(R--P--S,green); draw(circumcircle(C,L,N)^^circumcircle(B,K,M),dotted); dot("$X$",X); dot("$B$",B,dir(90)); dot("$C$",C,dir(-130)); dot("$M$",M,dir(90)); dot("$N$",N,dir(N)); dot("$K$",K,dir(20)); dot("$L$",L,dir(20)); dot("$P$",P,dir(180)); dot("$R$",R); dot("$S$",S,dir(60)); dot("$T$",T,dir(T)); [/asy][/asy] Claim 3: $KL,BC,MN$ concur at some point $T$. Proof: All the three points $B,C, KL \cap MN$ lie on polar of $MK \cap NL$ wrt $\Gamma$. $\square$ Claim 4: $ML,NK$ concur at $R$. Proof: Let $\infty$ be point at infinity along lines $BM,CN,PR$. Then $\Gamma$ is incircle of (degenerate) hexagon $$ PBM \infty CL $$By Brianchon, it follows that $$ P\infty, BC,ML $$concur. As $P\infty \cap BC = R$, hence $R \in ML$. Similarly $R \in NK$. $\square$ Claim 5: $M,N,R,S$ are cyclic. Proof: We angle chase. Note $XNCLS,XMBKS$ are cyclic on $\odot(XC),\odot(XB)$, respectively. Thus, \begin{align*} \angle MSN &= 180^\circ - (\angle NSC + \angle MSB) = 180^\circ - (\angle NLC + \angle MKB) \\ &= 180^\circ - (\angle NML + \angle MNK) = 180^\circ - (\angle RMN + \angle RNM) = \angle MRN \end{align*}proving our Claim. $\square$ Corollary 6: $R,S,K,L$ are cyclic. Proof: By PoP we obtain $$TK \cdot TL = TM \cdot TN = TR \cdot TS$$as desired. $\square$ Claim 7: $P$ is circumcenter of $\triangle RKL$. Proof: We will show $PL = PR$. Let $P_0 = RL \cap CN$. Then $\angle P_0LN = 90^\circ$. Also, $CL = CN$. It follows $C$ is circumcenter of $\triangle P_0LN$. Thus $CL=CP_0$. Since $$\triangle CLP_0 \sim \triangle PLR$$so it follows $PL = PR$, as desired. $\square$ It follows that points $K,L,R,S$ lie on a circle with center $P$. Thus $$ \boxed{PR = PS} $$which completes the proof! $\blacksquare$

This problem was proposed by Burii.

We can first get rid of $R$ by noting that $PQ=PR$ by Ceva's. This was probably a bad idea but whatever. We can start with a little bit of length bashing. $ABPQ$ circumscribed means that $AB+PQ=BP+AQ$. $AB=QP\cdot \frac{AQ+QD}{QD}$ and $BP=DP\cdot \frac{AQ}{QD}$. Thus $DP\cdot \frac{AQ}{QD}+AQ=QP+QP\cdot \frac{AQ+QD}{QD},$ or $$DP=QP\cdot \frac{2\cdot QD+AQ}{AQ}-QD.$$Let $AQ=x,QD=y,QP=z$. Then $DP=z\cdot \frac{2y+x}{x}-y$ and by symmetry $AP=z\cdot \frac{2x+y}{y}-x$. Stewart's theorem in $\triangle APD$ gives us that $(x+y)xy+z^2(x+y)=\left(z\cdot \frac{2x+y}{y}-x\right)^2\cdot y+\left(z\cdot \frac{2y+x}{x}-y\right)^2\cdot x$. This turns into $z^2\cdot \left(\frac{4x^2+4xy}{y}+\frac{4y^2+4xy}{x}\right)=z\cdot (4x^2+4xy+4y^2)$ or $$z=\frac{xy(x^2+xy+y^2)}{(x+y)(x^2+y^2)}.$$With this, $AP=\frac{x^2(x^2+2xy+2y^2)}{(x+y)(x^2+y^2)}$ and $DP=\frac{y^2(y^2+2xy+2x^2)}{(x+y)(x^2+y^2)}$. Now we can do geo. We can get $AC=AP\cdot \frac{AD}{AQ}$ and $BD=DP\cdot \frac{AD}{DQ}$, so using the above (we didn't actually need Stewart's for this part), $AB+AC=BD+CD$, and there is a circle $\Gamma$ tangent to $\overline{AB},\overline{AC},\overline{BD},\overline{CD}$ centered at $X$. We can also get from the above bash, with enough work, that $\frac{AB+BP-AP}{2}=QP$. This means that the tangents from $P$ to the $A$-excircle of $\triangle ABP$, or $\Gamma$, have length $PQ$. Let the tangency point to $\Gamma$ on $\overline{AC}$ be $Y$ and the tangency point on $\overline{AB}$ be $Z$. Then $PQ=PY$, so $\overline{QY}$ is parallel to the angle bisector of $\angle QPD$, and $BY=BZ$, so $\overline{YZ}$ is parallel to the angle bisector of $\angle ABD$. This means that $Q-Y-Z$, or that $Q$ is on the polar of $B$ with respect to $\Gamma$. By symmetry, $Q$ is on the polar of $C$ with respect to $\Gamma$, or $Q$ is the pole of $\overline{BC}$ with respect to $\Gamma$. $S$ is then the inverse of $Q$ with respect to $\Gamma$, but because $(P,PQ)$ is orthogonal to $\Gamma$, $S$ is on $(P,PQ)$, as desired.