P-isosceles $\triangle DPA \sim \triangle BPC$ are similar. Let points $B_0 \in (PB, C_0 \in (PC$ be such that $\triangle DPA \cong \triangle B_0PC_0$ are congruent $\Longrightarrow$ $B_0C_0 \parallel BC$. $P$ is circumcenter of $AB_0C_0D$ with $DA = B_0C_0$ $\Longrightarrow$ it is isosceles trapezoid with $AB_0 \parallel DC_0$. Let $Q_0 \equiv AC_0 \cap B_0D$ be its diagonal intersection. Let $S$ be circumcenter of $\triangle DPA$. Undefine $Q$. $S$-isosceles $\triangle ASP \cong \triangle PSD$ and $Q_0$-isosceles $\triangle AQ_0B_0 \sim \triangle C_0Q_0D$ are all similar $\Longrightarrow$ segment $|SQ_0|$ is spirally similar to segments $|PB_0|, |PC_0|$ with centers $A, D$ and the same coefficient $\frac{AQ_0}{AB_0} = \frac{AS}{AP} = k = \frac{DS}{DP} = \frac{DQ_0}{DC_0}$. Since $\frac{\overline{BP}}{\overline{BB_0}} = \frac{\overline{CP}}{\overline{CC_0}},$ these spiral similarities take $B \in PB_0, C \in PC_0$ to the same point $Q \in SQ_0$, such that $\angle QAB = \angle SAP = \angle PDS = \angle QDC$ and $\frac{AQ}{AB} = k = \frac{DQ}{DC}$ $\Longrightarrow$ $\triangle AQB \sim \triangle CQD$ are Q-isosceles and similar $\Longrightarrow$ $Q$ is intersection of perpendicular bisectors of $|AB|, |CD|$ and $\angle AQB = \angle CQD$.

If $\angle APD=\angle BPC,$ then $\angle APC=\angle DPB$ $\Longrightarrow$ $\triangle APC$ and $\triangle DPB$ are congruent by SAS criterion, since $PA=PD$ and $PC=PB.$ Therefore, $AC=BD,$ which implies that $\triangle AQC$ and $\triangle BQD$ are congruent by SSS criterion, due to $QA=QB$ and $QC=QD$ $\Longrightarrow$ $\angle AQC=\angle BQD$ $\Longrightarrow$ $\angle AQB=\angle CQD.$