Solution:

By the tangent-chord angle, $\angle DBC = \angle DCB = \angle A,$ so triangle $DBC$ is isosceles with $DB = \frac{BC/2}{\cos A}.$ The law of cosines gives $\cos A = \frac{10^2 + 5^2 - 9^2}{2 \cdot 10 \cdot 5} = \frac{11}{25}$ and $\cos B = \frac{9^2 + 5^2 - 10^2}{2 \cdot 9 \cdot 5} = \frac{1}{15},$ so $DB = \frac{9/2}{11/25} = \frac{225}{22}.$

Since $D$ lies on the opposite side of $BC$ from $A,$ $\angle ABD = A + B,$ and $$\cos(A + B) = \frac{11}{25} \cdot \frac{1}{15} - \frac{6\sqrt{14}}{25} \cdot \frac{4\sqrt{14}}{15} = \frac{11 - 336}{375} = -\frac{13}{15}.$$ The law of cosines in triangle $ABD$ then gives $$DA^2 = 5^2 + \left(\tfrac{225}{22}\right)^2 + 2 \cdot 5 \cdot \tfrac{225}{22} \cdot \tfrac{13}{15} = \frac{105625}{484}, \qquad DA = \frac{325}{22}.$$

The power of $D$ gives $DP \cdot DA = DB^2,$ so $$AP = DA - \frac{DB^2}{DA} = \frac{DA^2 - DB^2}{DA} = \frac{(105625 - 50625)/484}{325/22} = \frac{55000}{22 \cdot 325} = \frac{100}{13},$$ and $m + n = 100 + 13 = 113.$