i) AB & PQ are both chords of first circle intersect at E.
So by two chords intersecting theorem of a circle, PE*EQ = AE*EB ------ (1)
ii) Similarly, from circle (2), we have, RE*ES = AE*EB -------- (2)
iii) Thus from (1) & (2): PE*EQ = RE*ES
==> PE/ES = RE/EQ
Also, <PES = <REQ [Vertically opposite angles]
So ΔPES is similar to ΔREQ [SAS similarity axiom]
==> <SPE = <QRE
As these two are angles of the same segment SQ and being equal,
Points P,R,Q,S lie on the same circle,
So PRQS is a cyclic quadrilateral.