Correct Answer: $36^{\circ}$

Solution :

Point D is taken on the major arc of the circle. Then, AD, BD, CB, and AC are joined.$\angle ACB = 108^\circ$
Now,
$\angle ADB = (180 - 108)^\circ = 72^\circ$
Now,
$\angle AOB = (72\times 2)^\circ = 144^\circ$
Since $OA = OB,\triangle AOB$ is an isosceles triangle and $\angle OBA = \angle OAB$.
Now,
$\angle AOB + \angle OBA + \angle OAB = 180^\circ$
⇒ $2 \times \angle OBA = (180 - 144)^\circ$
⇒ $\angle OBA = 18^\circ$
Now,
$\angle OBA = \angle OAB = 18^\circ$
Now,
$\angle OAP = \angle OAB + \angle BAP$
⇒ $90^\circ = 18^\circ + \angle BAP$
⇒ $\angle BAP = 72^\circ$
Now,
$\angle OBP = \angle OBA + \angle ABP$
⇒ $90^\circ = 18^\circ + \angle ABP$
⇒ $\angle ABP = 72^\circ$
Now,
$\angle APB + \angle ABP + \angle BAP = 180^\circ$
⇒ $\angle APB + 72^\circ + 72^\circ = 180^\circ$
⇒ $\angle APB = 36^\circ$
Hence, the correct answer is $36^\circ$.