Let's assume $W$ is the midpoint of arc $BC$. As is known, points $W$, $M$, and $N$ are collinear, and $W$ is the midpoint of the circumscribed circle of triangle $BIC$ (see fig. 47). Also, it is known that the center of the excircle of triangle $ABC$ that touches $BC$ belongs to this circle. Let $IK$ cross $TN$ at point $Q$. Since $\angle IAT = \angle TAW = \angle TNW = \angle TQI$, the quadrilateral $AITQ$ is inscribed. Then, according to the theorem about the multiplication of the line segments of chords, $IK \cdot KQ = AK \cdot KT = BK \cdot KC$, so the quadrilateral $IBQC$ is also inscribed, i.e., $Q$ belongs to the circumscribed circle of triangle $BIC$. $\angle NBW = \angle NCW = 90^\circ$, so $NB$ and $NC$ are tangent lines to the circumscribed circle of triangle $BIC$. In triangle $BPC$, the line $PN$ is the symmedian, so $\angle BPQ = \angle CPM$. Let's assume $PM$ crosses the circumscribed circle of triangle $BIC$ a second time at point $I_a$. Let's prove that $I_a$ is the center of the excircle of triangle $ABC$. As already noted, $\angle BPQ = \angle CPI_a$, so the arcs $BQ$ and $CI_a$ of the circumscribed circle of triangle $BIC$ are equal. Consider the symmetry relative to $NW$. Point $B$ maps to point $C$, and point $Q$ maps to point $I_a$, since the arcs $BQ$ and $CI_a$ are symmetrical relative to $NW$. Let the perpendicular from $I_a$ to $BC$ cross $BC$ at point $X$, then according to symmetry the line segments $MK$ and $MX$ are equal, so $BK = CX$, i.e., $X$ is the touchpoint of the excircle to the $BC$ side. So, $I_a$ belongs to the circumscribed circle of triangle $BIC$, lies with $A$ in a different half-space relative to $NW$, and lies on the perpendicular that passes through the touchpoint of the excircle of triangle $ABC$.

Obviously, there is only one such point, and the center of the excircle of triangle $ABC$ that touches the $BC$ side satisfies such conditions. So, $I_a$ is the center of the excircle.

Thus, to finish the solution, it's enough to show that $AK \parallel PI_a$. This statement follows from the fact that $\angle AI_aP = \angle IQP = \angle TQI = \angle TAI = \angle KAI_a$.