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Since $△MPN$ is an isosceles triangle, we have $\angle PMA=\angle PMN=\angle MNP=\angle BNP$. By tangent and chord theorem, $\angle MCA=\angle PMA=\angle BNP=\angle BDN$. Since $\angle MCP=\angle MNP$, the quadrilateral $CMPN$ is cyclic. Analogously, from $\angle PDN=\angle PMN$, we get that $NDMP$ is cyclic. Since $C$ and $D$ both lie on the circumcircle of $△NPM$, points $P,N,M,C$ and $D$ are concyclic. From inscribed angles subtending arcs with the same length, we get that $\angle MDP=\angle MCP=\angle MNP=\angle PDN=\angle PMN=\angle PCN$. The power of $P$ with respect to ${\omega }_1$ gives us that $|PM{|}^2=|PA|\cdot |PC|$. The power of $P$ with respect to ${\omega }_2$ gives us that $|PN{|}^2=|PB|\cdot |PD|$. Since $|PM|=|PN|$, the powers of $P$ with respect to ${\omega }_1$ and ${\omega }_2$ are equal ($P$ lies on the radical axis). Hence, $|PA|\cdot |PC|=|PB|\cdot |PD|$, which implies that $ABDC$ is cyclic. From inscribed angles subtending the arc $AB$, we get that $\angle ACB=\angle ADB$. Hence, $\angle BCN=\angle ACN-\angle ACB=\angle MDB-\angle ADB=\angle MDA$.

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Alternative solution.

Since $△MPN$ is an isosceles triangle, we have $\angle PMA=\angle PMN=\angle MNP=\angle BNP$. By tangent and chord theorem, $\angle MCA=\angle PMA=\angle BNP=\angle BDN$. Since $\angle MCP=\angle MNP$, the quadrilateral MPNC is cyclic, which means that $P$ lies on the circumcircle of $△MNC$. Since $△MPN$ is isosceles, the perpendicular bisector of $MN$ passes through $P$. Since the intersection point of the angle bisector and the perpendicular bisector of the opposite side of the triangle lies on the circumcircle, it follows that $CP$ bisects angle $\angle MCN$. Hence, $\angle MCP=\angle PCN$. Analogously, since $\angle PDN=\angle PMN$, it follows that $NDMP$ is cyclic and the circumcircle of $△MND$, the perpendicular bisector of $MN$ and the angle bisector of $\angle MDN$ meet at $P$. Hence, $\angle MDP=\angle PDN=\angle MCP=\angle PCN$. Now we continue as in the previous solution.