Solution:

The case $N = B$ shall be excluded in what follows. Then $DN$ is not orthogonal to $AC$, and $M$ is uniquely characterized as the intersection point of the perpendicular bisector of $DN$ with $AC$. In the triangle $DNC$, $AC$ is the angle bisector at $C$, and in every triangle the angle bisector and the perpendicular bisector of the opposite side intersect on the corresponding circumcircle arc, so $DMNC$ is a cyclic quadrilateral; in particular $\text{MDN = MCN}$. Now $\text{MDN = RDP}$ and $\text{MCN = ACB = BAC = RAP}$, so by the inscribed angle theorem $ARPD$ is also a cyclic quadrilateral. In its circumcircle, because of $\text{RAP = BAC = CAD = PAD}$, the chords $RP$ and $PD$ are of equal length (law of sines or inscribed angle theorem).