The measure of segment AB on the parallelogram $ABCD$ below?

Question

For reference
Through the vertex A of a parallelogram ABCD traces AP(P in BC) of
such a way that $\measuredangle BAP= 2m\measuredangle PAD.$. The height $BH$ intercepts $AP$ on "Q". Calculate $AB$ if $PQ$ = 20

My progress:

I drew the angles and some auxiliary lines but an equation is still missing

Answer 1

Let $O$ be the middle of $PQ$. Then $\angle BPO = \angle PBO = x,$ so $BO = 0.5PQ = 10$ and $\angle BOQ = 2x,$ and then $\Delta ABO$ is isosceles. Thus $AB = BO = 10.$