Solution:

Let $\omega_{1}$ and $\omega_{2}$ be the two incircles, centered at $I_{1}$ and $I_{2}$ respectively. We first introduce the points $R_{1}$ and $S_{1}$ on $\omega_{1}$ such that $XR_{1}$ and $XS_{1}$ are tangent to $\omega_{1}$, with the condition that $S_{1}$ is on the line $AP$. Similarly, we introduce the points $R_{2}$ and $S_{2}$ on $\omega_{2}$ such that $XR_{2}$ and $XS_{2}$ are tangent to $\omega_{2}$, with the condition that $S_{2}$ is on the line $AP$. Finally, let $T_{1}$ and $T_{2}$ be the contact points of $\omega_{1}$ and $\omega_{2}$ on the line $BC$. Noting that $I_{1}, I_{2}$ are on the angle bisectors of $\angle R_{1}XP, \angle PX R_{2}$, respectively, we find that
$$\angle R_{1}XR_{2} = \angle R_{1}XP + \angle PX R_{2} = 2 \cdot (\angle I_{1}XP + \angle PX I_{2}) = 2 \cdot 90^{\circ} = 180^{\circ}$$
so that $R_{1}R_{2}$ is a common tangent of $\omega_{1}$ and $\omega_{2}$. Note that, by reflection over $I_{1}I_{2}$, we have $R_{1}R_{2} = T_{1}T_{2}$. Moreover, as tangents from a point have the same length, we can observe that
$$R_{1}R_{2} = R_{1}X + XR_{2} = XS_{1} + XS_{2} = S_{2}S_{1} + 2 \cdot XS_{2}$$
and
$$T_{1}T_{2} = T_{1}P + PT_{2} = PS_{1} + PS_{2} = S_{2}S_{1} + 2 \cdot PS_{1}$$
so that $XS_{2} = PS_{1}$, and we can calculate
$$AX = AS_{2} - XS_{2} = AS_{2} - PS_{1}$$
However, these last lengths can be computed as distances from a vertex of a triangle to a contact point of its incircle. Hence,
$$AX = AS_{2} - PS_{1} = \frac{AB + AP - BP}{2} - \frac{AP + PC - AC}{2} = \frac{AB + AC - BC}{2}$$
which is independent from the choice of $P$.