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

 $$PA = PB$$ and the angle at $$P$$ is $$72\degree$$

The base angles are 

$$\angle PAB = \angle PBA = \dfrac{1}{2} (180-72) =54\degree$$ (Angle sum property of an isosceles triangle. 

As $$AC$$ is parallel to $$PB$$

$$\angle CAB = \angle ABP = 54\degree$$(Interior alternate angles)

Now, using the alternate segment theorem,

$$\angle ACB = \angle ABP = 54\degree$$

$$\implies \angle CBA = 180\degree - 54\degree - 54\degree = 72\degree$$

Now, ABCD forms a cyclic quadrilateral, which implies that opposite angles are supplementary.

$$ \angle CDA = 180\degree- \angle CBA = 180\degree- 72\degree = 108\degree$$