Answer: 77

Solution:

Let ray $AD$ meet the circumcircle of $\triangle ABC$ again at $E.$ Since $AD$ bisects angle $A,$ the point $E$ is the midpoint of arc $BC,$ so $E$ lies on the perpendicular bisector of $\overline{BC}$ and projects onto line $BC$ at $M.$ The projections of the collinear points $A,$ $P,$ $E$ onto line $BC$ are $H,$ $N,$ $M,$ and projection preserves ratios along a line; since $N$ is the midpoint of $\overline{HM},$ point $P$ is the midpoint of $\overline{AE}.$

Here $\angle B = 105^\circ,$ and $\angle CBE = \angle CAE = 15^\circ$ (both subtend arc $CE$), so $\angle ABE = 120^\circ.$ Also $\angle AEB = \angle ACB = 45^\circ$ (both subtend arc $AB$). The law of sines in $\triangle ABE$ gives $$\begin{aligned} AE &= AB \cdot \frac{\sin \angle ABE}{\sin \angle AEB} \\ &= 10 \cdot \frac{\sin 120^\circ}{\sin 45^\circ} \\ &= 5\sqrt{6}. \end{aligned}$$

Therefore $AP = \frac{1}{2} AE = \frac{5\sqrt{6}}{2},$ so $AP^2 = \frac{75}{2}$ and $m + n = 75 + 2 = 77.$