In the adjoining figure, PA and PB are tangents to the circle. $$AC$$ is parallel to $$PB$$. Then measure of $$\angle CDA$$ is

Since $$PA = PB$$ and the angle at $$P$$ is $$72^\circ$$, the base angles give $$\angle PAB = \angle PBA = 54^\circ$$. As $$AC$$ is parallel to $$PB$$, alternate angles give $$\angle CAB = \angle ABP = 54^\circ$$, and the tangent-chord angle at $$B$$ together with the parallel lines gives $$\angle ACB = 54^\circ$$ as well, so $$\angle ABC = 180^\circ - 54^\circ - 54^\circ = 72^\circ$$. Since $$ABCD$$ is cyclic, $$\angle CDA = 180^\circ - 72^\circ = 108^\circ$$.