Solution:

The answer is (B). Let $O$ be the center of $\omega$; by hypothesis we have $\widehat{EAD}=\widehat{FAD}$, so $D$ is the midpoint of arc $EF$ and in particular $OD$ passes through the midpoint of segment $EF$. It follows that $OD$ is the perpendicular bisector of $EF$, since it is perpendicular to it and passes through its midpoint. Now, since $PE$ and $PF$ are tangents, we have $PE=PF$ and hence $P$ also belongs to the perpendicular bisector of $EF$. This implies that $P, D, O$ are collinear. The triangle $PEO$ is right-angled at $E$ by construction, so we have $PO^2=(PD+DO)^2=(PD+4)^2=EO^2+PE^2=4^2+3^2$, from which we obtain $PD=1$.