IR-TST 2017 Problem 14

Let $P$ be a point in the interior of quadrilateral $ABCD$ such that: $$\angle BPC=2\angle BAC \ \ ,\ \ \angle PCA = \angle PAD \ \ ,\ \ \angle PDA=\angle PAC$$ Prove that: $$\angle PBD= \left | \angle BCA - \angle PCA \right |$$ Proposed by Ali Zamani