RU-SPB 2011 Problem 107

$ABCD$ - convex quadrilateral. $P$ is such point on $AC$ and inside $\triangle ABD$ , that $$\angle ACD+\angle BDP = \angle ACB+ \angle DBP = 90-\angle BAD$$ . Prove that $\angle BAD+ \angle BCD =90$ or $\angle BDA + \angle CAB = 90$