$$QP$$ parallel to $$ST$$ and $$QT$$ is a transversal.
$$\therefore$$ $$\angle PQR = \angle STU$$ .... Alternate interior angles
$$US$$ parallel to $$PR$$ and $$UR$$ is a transversal.
$$\therefore$$ $$\angle PRQ = \angle SUT$$ ....Alternate interior angles
Here, two angles are equal to the two angles of another triangle. Hence, the third one will also equal.
Also, by $$AA$$ test, $$\Delta PQR\sim \Delta STU$$
$$\therefore$$ $$\angle QPR = \angle TSU$$
Now, given that $$\angle P = 90^{o}$$
$$\therefore$$ $$\angle S = 90^{o}$$