<Cir-202-797>, hint

Line QB cut AP at R, O be the center of the circle.

Let <ORA=<OBA=x, Since triangles OPA and OQB are congruent, 

<AOB=180-2x=<POQ, OP=OQ, <OPQ=<OQP=x, and <OBA=x. 

Hence O,Q,S,B are concyclic and <OBQ=<OSQ=90.

Finally PS=QS (OPQ is isosceles, )