Solution:

A homothety (scaling) about $P$ takes triangle $ADP$ into $BCP$, since $AD$, $BC$ are parallel and $A, P, C; B, P, D$ are collinear. The ratio of homothety is thus $\sqrt{2}$. It follows that, if we rescale to put $[ADP]=1$, then $[ABP]=[CDP]=\sqrt{2}$, just by the ratios of lengths of bases. So $[ABCD]=3+2\sqrt{2}$, so $[ADP]/[ABCD]=1/(3+2\sqrt{2})$. Simplifying this, we get $3-2\sqrt{2}$.