IGO 2018 I5

Suppose that $ABCD$ is a parallelogram such that $\angle DAC = 90^o$ . Let $H$ be the foot of perpendicular from $A$ to $DC$ , also let $P$ be a point along the line $AC$ such that the line $PD$ is tangent to the circumcircle of the triangle $ABD$ . Prove that $\angle PBA = \angle DBH$ .

Proposed by Iman Maghsoudi