Since we're given that $ABCD$ is cyclic, we've $$PS = AO \cdot \sin \angle BAD = AO \cdot \frac{BD}{2R},$$ where $R$ is the circumradius of $ABCD$. Analogously $$QR = OC \cdot \sin \angle BCD = CO \cdot \frac{BD}{BR}.$$ Summing yields $$PS + QR = \frac{AC \cdot BD}{2R} \le BD,$$where we used $AC \le 2R$. $\square$

First, it is well-known that $PQRS$ is circumscribed,so $SP+QR=PQ+SR$. Note that $BO\ge PQ$, because $PQ$ is a chord in a circle with diameter $BO$, also $DO\ge SR$,so $BD\ge PQ+SR$. $\square$