Let $ABCD$ be a convex quadrilateral with $\angle DAB =\angle ABC =\angle BCD > 90^o$. The circle circumscribed around the triangle $ABC$ intersects the sides $AD$ and $CD$ at points $K$ and $L$, respectively, different from any vertex of the quadrilateral $ABCD$ . Segments $AL$ and $CK$ intersect at point $P$. Prove that $\angle ADB =\angle PDC$.