IMO 2020 Problem 1

Consider the convex quadrilateral $ABCD$ . The point $P$ is in the interior of $ABCD$ . The following ratio equalities hold: $$\angle PAD:\angle PBA:\angle DPA=1:2:3=\angle CBP:\angle BAP:\angle BPC$$ Prove that the following three lines meet in a point: the internal bisectors of angles $\angle ADP$ and $\angle PCB$ and the perpendicular bisector of segment $AB$ .

Proposed by Dominik Burek, Poland