$\angle TRQ={{30}^{\circ }}$ .
At point R, OR ⊥ RQ.
So, $\angle ORQ={{90}^{\circ }}$
$\Rightarrow \angle TRQ+\angle ORT={{90}^{\circ }}$
$\Rightarrow \angle ORT={{90}^{\circ }}-{{30}^{\circ }}={{60}^{\circ }}$
It’s seen that, ST is the diameter,
So, $\angle SRT={{90}^{\circ }}$ [ ∵ Angle in semicircle = 90°]
Then,
$\angle ORT+\angle SRO={{90}^{\circ }}$
$\angle SRO+\angle PRS={{90}^{\circ }}$
$\therefore \angle PRS={{90}^{\circ }}-{{30}^{\circ }}={{60}^{\circ }}$