First, let $E$ be the reflection of $D$ over $AC$; then $A$ is the circumcenter of $(EPD)$. Claim 1: Let $K = DP \cap (ABCD)$; then $K$ is the circumcenter of $(PBQE)$. Proof: $\angle{EAP} = 2\angle{EDP} \implies K$ is the midpoint of $\widehat{EB}$, so $KE = KB$, and $\angle{EKP} = 2\angle{EBP}$ so $K$ is the circumcenter of $(PQ)$. Also, $\triangle{EPQ} \sim \triangle{EAC} \implies \triangle{EAP} \sim \triangle{ECQ} \implies CE = CQ = CD$. Let $Q'$ be the $Q$-antipode wrt $(QED)$; then $E, P, Q'$ are collinear, and $\triangle{PED} \sim \triangle{PQQ'}$. Now if $Q''$ is such that $\triangle{QED} \sim \triangle{Q''QQ'}$, then $Q''$ is on $PQ'$ such that $QQ'' \perp QD$, and $\angle{PSQ} = \angle{PCQ''}$, so it suffices to show that $C, R, Q''$ are collinear. Claim 2: If $X = CR \cap (ABCD)$, then $Q''QXE$ is cyclic. Proof: $\angle{EQ''Q} = \angle{EQ'D} = \angle{ECA} = \angle{EXA}$. Thus, $\angle{CXQ} = \angle{Q''XQ} = 90^{\circ} \implies C, R, X, Q''$ are collinear, as desired. $\square$

this problem ruined my geo exam perfect score