<Cir-2024-768>, hint

MB=MC ( radius), MQ is common, <BQM=<CQM=90.

Thus triangles MBQ and MCQ are congruent and Q is the midpoint of BC.

Hence P,Q,R,S are the midpoints of  quadrilateral ABCD and hence PQ//SR, PS//QR

and AC and BD are perpendicular,

Hence angles P,Q,R,s are right angle and so PQRS is rectangle.