Find the power of $P$ in relation the center circumference $O$ below.

$ - 5$ is not the correct answer.

Note that $BH = 2 OD = 4$ and hence $BP = \sqrt{15}$

We then use the fact that $\triangle ACP \sim \triangle BHP$

$ \displaystyle \implies AC = \frac{4}{\sqrt{15}} (AH+1), ~ PC = \frac{1}{\sqrt{15}} (AH+1)$

If $M$ is the midpoint of $BC$, $ \displaystyle CM = \frac {\sqrt{15} + PC}{2}$

$ \displaystyle PM = CM - PC = \frac {\sqrt{15} - PC}{2} = \frac{14 - AH}{2 \sqrt{15}}$

Power of point $P$ with respect to the circumcenter $O = OP^2 - OA^2$,

$ \displaystyle = OM^2 + PM^2 - (AD^2 + 4)$

$ \displaystyle = \frac{AH^2}{4} + \frac{(14-AH)^2}{60} - \frac{AC^2}{4} - 4$

$ \displaystyle = \frac{AH^2}{4} + \frac{(14-AH)^2}{60} - \frac{4 (AH+1)^2}{15} - 4 = - AH - 1$

So it depends on $AH$ and unless $\triangle ABC$ is isosceles with $\angle A = \angle B$, the answer cannot be $ - 5$.