Shift point $P$ towards right to position $P'$ such that $PP' = AD$ and $PP' \parallel AD$, as shown.

We find that both $ADP'P$ and $BPP'C$ are parallelograms. Therefore $$\angle{P'PC} = \angle{PCB} = \angle{PAB}=\angle{P'DC}$$ \indent This means that points $P, C, P',$ and $D$ are concyclic. It follows that: $$\angle{PBA}=\angle{P'CD}=\angle{P'PD}=\angle{PDA}$$