Let $Q'$ be the intersection of the line passing $A$ and parallel to $BC$ and the line passing $P$ and perpendicular to $AD$. Let $U$ be the intersection of $DI$ and $AQ'$, and $V$ be the intersection of $PQ'$ and the circle $I$. ($V \neq P$)
Since $\angle VPD = 90^\circ$, four points $D, I, V, U$ lie on a line.
Since $\angle BDI = 90^\circ$ we have $\angle AUI = 90^\circ$ and since $\angle AFI = \angle AEI = 90^\circ$ we have that five points $A, F, I, E, U$ lie on a circle, say, $C_1$.
Four points $A, P, V, U$ lie on a circle, say, $C_2$, because $\angle APV = \angle AUV = 90^\circ$.
For given three circles, three perpendicular bisectors of the line segments joining two centers of two circles meet at one point. Considering three circles $C_1, C_2$, and the incircle $I$, we have that three lines $AU, PV, EF$ meet at one point $Q'$. Thus we have $Q' = Q$.
Since $\triangle AXE \sim \triangle CDE$, we have $\frac{AX}{AE} = \frac{CD}{CE}$. So
$$AX = \frac{CD}{CE} \times AE = AE.$$
Since $\triangle AYF \sim \triangle BDF$, we have $\frac{AY}{AF} = \frac{BD}{BF}$. So
$$AY = \frac{BD}{BF} \times AF = AF.$$
Since $AE = AF$ we have $AX = AY$, which completes the proof. $\square$