Let $H$ be the orthocenter of triangle $D E F$. We claim that $P$ is the midpoint of $\overline{D H}$. Indeed, consider an inversion at the incircle of $A B C$, denoting the inverse of a point with an asterik. It maps $A B C$ to the nine-point circle of $\triangle D E F$. According to $\angle I A X=90^{\circ}$, we have $\angle A^{*} X^{*} I=90^{\circ}$. Hence line $X I$ passes through the point diametrically opposite to $A^{*}$, which is the midpoint of $\overline{D H}$, as claimed. The rest is a straightforward computation. The inradius of $\triangle A B C$ is $r=4$. The length of $E F$ is given by $E F=2 \frac{A F \cdot r}{A I}=\frac{16}{\sqrt{5}}$. Then, $D P^{2}=\left(\frac{1}{2} D H\right)^{2}=\frac{1}{4}\left(4 r^{2}-E F^{2}\right)=4^{2}-\frac{64}{5}=\frac{16}{5}$. Hence $D P=\frac{4 \sqrt{5}}{5}$.