Since $SP = SQ$, the triangle $PSQ$ is isosceles. Therefore, $\angle SPQ = \angle PQS$. We denote the measure of those angles by $y$. Similarly, $\angle SQR = \angle QRS = x^\circ$.
Since the sums of the interior angles of $PQR$ is $180^\circ$, $x + y + (x+y) = 180$, so $ 2x+2y=180 $.
Therefore, $ \angle PQR = x^\circ+y^\circ=90^\circ $.
This problem is taken from the UKMT Mathematical Challenges.