Correct Answer: $150^{\circ}$

Solution :
Chord AB = Radius
In $\triangle$OAB,
OA = BO = Radius
⇒ $\triangle$OAB is an equilateral triangle.
⇒ $\angle$AOB = $60^{\circ}$
Since the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
⇒ $\angle$AOB = 2 × $\angle$ACB
⇒ $\angle$ACB = $\frac{60^{\circ}}{2}$ = $30^{\circ}$
Since ACBD is a cyclic quadrilateral and opposite angles of a cyclic quadrilateral are supplementary,
$\angle$ACB + $\angle$ADB = $180^{\circ}$
⇒ $\angle$ADB = $180^{\circ}- 30^{\circ}=150^{\circ}$
Hence, the correct answer is $150^{\circ}$.