Clearly, $M$ is a point on $BC$, since $I_a B$ is angle bisector of both angles $\overline{XBM}$ and $\overline{XBC}$. Note that
$$\overline{I_a MC} = 180^\circ - \overline{I_a MB} = 180^\circ - \overline{I_a XA} = \overline{CYI_a}.$$
And since $\overline{I_a CM} = \overline{I_a CY}$, we get $CYI_a \equiv CMI_a$. Therefore $M$ is also the reflection of $Y$ over $I_a C$. For quadrilateral $KSI_a T$ we have
$$\overline{KSI_a} + \overline{KTI_a} = \overline{AYI_a} + \overline{AXI_a} = 180^\circ.$$
Therefore, $KSI_a T$ is cyclic. Now since $\overline{BMI_a} = \overline{BXI_a} = \overline{BTI_a}$, we get $\overline{BMTI_a}$ and similarly, $\overline{CMSI_a}$ are cyclic. This means $M$ is the Miquel point of the complete quadrilateral $KSI_a T$.

Let $\omega_1$ and $\omega_2$ be the circumcircles of triangles $CSI_a$ and $BTI_a$, respectively. The point $M$ has the same Simson line with respect to each of these two triangles, and since $X$ and $Y$ are the reflections of $M$ in $I_aB$ and $I_aC$, respectively, line $XY$ is homothetic with the Simson line of $M$, with center $M$ and radius 2. So the problem is to prove that the Simson line of $M$ with respect to $\omega_1$ and $\omega_2$ passes through the midpoint of $MZ$.

It is well-known that the Simson line of any point $P$ lying on the circumcircle of some triangle $QRU$, passes through the midpoint of $PH$ where $H$ is the orthocenter of $QRU$. So according to this fact, the Simson line of $M$ passes through the midpoint of $MH_1$ and $MH_2$ where $H_1$ and $H_2$ are orthocenters of triangles $BTI_a$ and $CSI_a$, respectively. Now, it is sufficient to show that $Z$ lies on $H_1H_2$. Consider two circles $\Gamma$ and $\Omega$ with diameters $ST$ and $KI_a$, respectively. Let $T'$ be the foot of the perpendicular line from $T$ to $BI_a$. We know that $T' \in \Gamma$. Also let $I'_a$ be the foot of the perpendicular line from $I_a$ to $BT$. Similarly $I'_a \in \Omega$. Note that $Z$ is a point on the radical axis of $\Gamma$ and $\Omega$, and since $I_aI'_a \cap TT' = H_1$, we get $\mathcal{P}_{\Gamma}(H_1) = H_1T \cdot H_1T' = HI_a \cdot HI'_a = \mathcal{P}_{\Omega}(H_1)$, where $\mathcal{P}_{\lambda}(Q)$ is the power of point $Q$ with respect to circle $\lambda$. Therefore $H_1$ and similarly $H_2$ lie on the radical axis of $\Gamma$ and $\Omega$. So the line passing through $Z$, $H_1$ and $H_2$, is the radical axis of $\Gamma$ and $\Omega$ and hence the claim of the problem. ■