Let $D$ be the foot of altitude from $A$. According to the statement of the problem, $ASDB$ is cyclic. Then $\angle SDC = \angle SAB = \angle APS$. Therefore, $CSDP$ would also be cyclic. Let $BC$ meet the circumcircle of $AHD$ for the second time at $E$. Since $\angle AEC = \angle AHC = 180^\circ - \angle B$. Thus, $AB = AE$ and $D$ is the midpoint of $BE$. That is, $\angle SED = \angle SAP = \frac{\angle CS}{2}$, $\angle SDE = \angle SPA$. Hence, $\triangle SED \sim \triangle SPA$. That is, $SD/SP = DE/PA$.

Let the circle with diameter $AC$ and the circumcircle of $BCP$ meet for the second time at $Q \neq C$, it similarly follows that $\angle QBD = \angle QPA = \frac{\angle CQ}{2}$, $\angle QDB = \angle QAP = \frac{\angle CQ}{2}$. Hence, $\triangle QBD \sim \triangle QPA$. That is, $DB/PA = QB/QP$. Since $DE = DB$ and (12) and (12), it follows that $QB/QP = SD/SP$. Taking into account that $BQPD, CSPD$ are cyclic, it follows that $\angle BQP = \angle DSP = 180^\circ - \angle C$. Further, $\triangle DSP \sim \triangle BQP$. Then

$$\angle SCB = \angle SPD = \angle QPB = \angle QCB.$$

Thus, $Q, S, C$ would be collinear. This completes our proof and we are finally done. ■