smc

Let $L$, $M$, $N$, and $O$ be the intersections of $AC$ and $PS$, $BD$ and $PQ$, $CA$ and $QR$, and $DB$ and $RS$ respectively. Since the circumcenter of a triangle is determined by the intersection of its perpendicular bisectors, each of the pairs mentioned earlier are perpendicular to each other. Let $\angle SPQ=\alpha$. Then $\angle LEM=180^{\circ }-\alpha$ since $\angle PLC=PMD=90^{\circ }$. This means $\angle DEC=180^{\circ }-\alpha$, which similarly implies $\angle QRS=\alpha$. Through a similar method, we find that $\angle PSR=\angle PQR=180^{\circ }-\alpha$. Since the opposite angles of $PQRS$ are congruent, $PQRS$ is indeed a parallelogram - and we showed this without any assumption about the type of quadrilateral $ABCD$ was. Therefore, $PQRS$ is always a parallelogram, and so our answer is $\boxed{PQRS\text{ is a parallelogram}}$.