Let $T = QR \cap AC$ and $V = QS \cap AB$. First, note that from $AP = AD$, we have $\angle APD = \angle ADP$. So
$$\angle CQP = 90^\circ - \angle QPB = 90^\circ - \angle APD = 90^\circ - \angle ADP = \angle QDC,$$
and hence $CQ = CD$. Now, by metric relations in the right triangle $DAC$, we have $AD^2 = AS \cdot AC$, then $AP^2 = AS \cdot AC$ and thus $\triangle APS \sim \triangle ACP$, where $\angle APS = \angle ACP$. In a similar fashion we get $CQ^2 = CD^2 = CS \cdot CA$, which implies $\angle CQS = \angle CAQ$.
Observe that $QATB$ is a cyclic quadrilateral because $\angle QBA = \angle QTA = 90^\circ$. Hence $\angle CQS = \angle CAQ = \angle CBT$, i.e., $SQ \parallel TB$. Therefore, $\angle AVS = \angle ABT$. But $\angle AVS = \angle APS + \angle PSQ$, and $\angle ABT = \angle RBT = \angle RCT$ (cyclic $RBCT$). Putting all together we get
$$\angle APS + \angle PSQ = \angle RCT = \angle RCP + \angle ACP \Rightarrow \angle PSQ = \angle RCP.$$