Given PA and PB are tangents to the circle with centre O from an external point P.

We know that the tangent at any point of a circle is perpendicular to radius at the point of contact.

$\therefore PA\perp OA,i.e.,\angle OAP={90}^{\circ }$      . .  . (i)
and  $PB\perp OB,i.e.,\angle OBP={90}^{\circ }$         . .. (ii)

Now, the sum of all the angles of a quadrilateral is ${360}^{\circ }$

$\therefore \angle AOB+\angle OAP+\angle APB+\angle OBP={360}^{\circ }$
$\Rightarrow \angle AOB+\angle APB={180}^{\circ }$  [using (i) and (ii)]

$\therefore$   quadrilateral OAPB is cyclic   [since both pairs of opposite angles have the sum ${180}^{\circ }$].