The points $D, E, M, N$ are concyclic if and only if $\angle DEP = \angle BMD$. Read from $\omega$ to infer that $\angle DEP = \angle DTP = \angle ADP$.
We now show that $\angle DTP = \angle ATE$. To this end, consider the homothety from $T$ mapping $\omega$ onto $\Omega$. This homothety maps $D$ to the midpoint $C'$ of the arc $AB$ not containing $C$, and $E$ to the midpoint $B'$ of the arc $AC$ not containing $B$. (A tangent to $\omega$ is mapped onto a parallel tangent to $\Omega$.)
Apply Pascal's theorem to the hexagram $ABB'TC'C$ to infer that $D, I, E$ are collinear, so $I$ is the midpoint of the segment $DE$ (Veldkamp, 1976-1977).
Since $AD$ and $AE$ are the tangents from $A$ to $\omega$, the line $AT$ is the $T$-symmedian in the triangle $DET$, so $\angle DTP = \angle DTI = \angle ATE$.
Next, $\angle ATE = \angle ATB' = \angle ABB'$ (read from $\Omega$), so $\angle ADP = \angle DTP = \angle ATE = \angle ABB' = \angle ABI$, showing that the line through $D, M, P$ is parallel to the internal bisectrix $BI$ of the angle $ABC$.
Consequently, $\angle BMD = \angle CBI = \angle ABI = \angle ADP = \angle DEP$, so the points $D, E, M, N$ lie on a circle $\gamma$.
Finally, we show that $\gamma$ is centered at the $A$-excenter of the triangle $ABC$. The center of $\gamma$ is the point where the perpendicular bisectrix of the segment $DE$ crosses the perpendicular bisectrix of the segment $DM$. The former is the internal $A$-bisectrix in the triangle $ABC$. The latter is the external $B$-bisectrix in the triangle $ABC$, since $\angle BDM = \angle ADP = \angle BMD$, so the triangle $BDM$ is isosceles with apex at $B$. The two angle bisectors cross at the $A$-excenter of the triangle $ABC$ and the conclusion follows.