The correct option is A 140
We know that, tangents drawn at any point on the circle is perpendicular to the radius through the point of contact.
∴OA⊥PA and OB⊥PB⇒∠OAP=∠OBP=90∘
Also, the sum of all four angles of quadrilateral is 360∘.
∴∠OAP+∠APB+∠OBP+∠AOB=360∘⇒90∘+40∘+90∘+∠AOB=360∘⇒∠AOB=360∘−90∘−40∘−90∘∴∠AOB=140∘