Answer: 578

Solution:

The hypotenuse is $AB = \sqrt{7^2 + 24^2} = 25,$ and the median to the hypotenuse gives $CM = \frac{25}{2}.$ Since $AD = BD,$ point $D$ lies on the perpendicular to $AB$ at $M,$ so $DM \perp AB$ and $$\begin{aligned} DM &= \sqrt{15^2 - \left(\tfrac{25}{2}\right)^2} \\ &= \sqrt{\tfrac{275}{4}} = \tfrac{5\sqrt{11}}{2}. \end{aligned}$$

Let $\beta = \angle AMC.$ In triangle $AMC$ with $AM = CM = \frac{25}{2}$ and $AC = 7,$ the law of cosines gives $$\begin{aligned} \cos\beta &= \frac{\left(\frac{25}{2}\right)^2 + \left(\frac{25}{2}\right)^2 - 7^2} {2 \cdot \frac{25}{2} \cdot \frac{25}{2}} \\ &= \frac{527}{625}. \end{aligned}$$ Since $C$ and $D$ are on the same side of $AB$ and $MD \perp AB,$ we have $\angle CMD = 90^\circ - \beta,$ so $\sin\angle CMD = \cos\beta.$

Therefore $$\begin{aligned} [CDM] &= \frac{1}{2} \cdot \frac{25}{2} \cdot \frac{5\sqrt{11}}{2} \cdot \frac{527}{625} \\ &= \frac{527\sqrt{11}}{40}, \end{aligned}$$ and $m + n + p = 527 + 11 + 40$ $= 578.$