Whether or not $P$ lies between the parallel lines gives two cases to distinguish. Let $S$ and $T$ be points on the tangent to the circumcircle of $\triangle ABP$ at $P$ such that $S$ and $A$ are on the same side of the line $BC$ and $T$ and $B$ are on the same side of $AD$.

First Solution:
We have $\angle SPA = \angle PBA$ (chord tangent angle). Because $AB \parallel CD$, we also have $\angle PBA = \angle PCD$, hence $\angle SPA = \angle PCD$. If $P$ lies between the parallel lines, $\angle SPA = \angle TPD$ and so $\angle PCD = \angle TPD$. In both cases it follows that $ST$ is tangent to the circumcircle of $\triangle DCP$.

Second Solution:
If $P$ is between the parallel lines, reflect $A$ and $B$ at $P$ to get $A'$ on $PD$ and $B'$ on $PC$. If $P$ is not between the parallel lines we simply let $A' = A$ and $B' = B$.

Because $A'B' \parallel CD$ the triangle $PDC$ is obtained from $\triangle PA'B'$ by a homothety with centre $P$. Hence, the circumcentres of the triangles $PDC$ and $PA'B'$ are on a line through $P$, on which we also find the circumcentre of $\triangle PAB$. As the tangent at $P$ to these circles is perpendicular to the line through $P$ and the centres, it follows now that the circumcircles of $\triangle PAB$ and $\triangle PDC$ touch at $P$.