Solution:

Let the diagonals of $PQRS$ intersect at $T$. Because $\overline{AP}$ and $\overline{AS}$ are tangent to $\omega$ at $P$ and $S$, we may write $\alpha = \angle ASP = \angle SPA = \angle SQP$ and $\beta = \angle CQR = \angle QRC = \angle QPR$. Then $\angle PTQ = \pi - \alpha - \beta$. On the other hand, $\angle PAS = \pi - 2\alpha$ and $\angle RCQ = \pi - 2\beta$, so that $ABCD$ is cyclic if and only if
$$\pi = \angle BAD + \angle DCB = 2\pi - 2\alpha - 2\beta$$
or simply
$$\pi/2 = \pi - \alpha - \beta = \angle PTQ$$
as desired.