Iranian Mathematical Olympiad

It is obvious that the bisector of $\angle B$ is perpendicular to $TK$ and so to the $AM$ (by assumption). Thus $BM = BA$, and the circumcenter of $AMC$ is on the bisector of $\angle B$ (the perpendicular bisector of $AM$). Since the center of $\omega$ is on the bisector of $\angle B$, and $P$ is on the line connecting the centers of these two circles, it is on the bisector of $\angle B$, too. Hence $\angle BAP = \angle BMP = 180^\circ - \angle PMC = \angle PAC$. So $P$ is on the bisector of $\angle A$ and hence $P$ is the incenter of $ABC$. So we have:
$$\angle TPK = \angle BTK = 90^\circ - \frac{\angle B}{2} = \frac{\angle A}{2} + \frac{\angle C}{2} = \angle TAP + \angle KCP$$
So if we draw the ray $Px$ such that $\angle TAP = \angle TPx$ and $\angle KCP = \angle KPx$, then this ray is tangent to the circumcircles of $ATP$ and $KPC$, and according to the above equation such ray exists. Thus the circles $APT$ and $KPC$ are tangent in $P$.