Correct Answer: 65

Solution :

Given, $\angle PQR=100^{\circ}$ and $\angle STR=105^{\circ}$
Since the angle subtended by the chord at the centre is double the angle subtended by the chord at the circumference,
reflex$\angle SOR = 2\angle STR = 210^\circ$
So, $\angle SOR = 360^\circ-210^\circ = 150^\circ$
By angle sum property in $\triangle SOR$,
$\angle SOR + \angle OSR+ \angle ORS =180^\circ$
⇒ $150^\circ+ 2\angle OSR =180^\circ$
⇒ $\angle OSR = 15^\circ$
Since PQRS is a cyclic quadrilateral,
$\angle PSR=180^\circ - \angle PQR$
⇒ $\angle PSR=180^\circ - 100^\circ $
⇒ $\angle PSR=80^\circ$
$\angle OSP = \angle PSR - \angle OSR$
⇒ $\angle OSP = 80^\circ - 15^\circ$
⇒ $\angle OSP = 65^\circ$
Hence, the correct answer is 65.