jmc

First, by the Law of Cosines, we have$$\cos BAC=\frac{16^2+10^2-14^2}{2\cdot 10\cdot 16}=\frac{256+100-196}{320}=\frac{1}{2},$$so $\angle BAC=60^{\circ }$. Let $O_1$ and $O_2$ be the circumcenters of triangles $B{I}_{B}D$ and $C{I}_{C}D$, respectively. We first compute$$\angle B{O}_{1}D=\angle B{O}_1{I}_B+\angle I_B{O}_{1}D=2\angle BD{I}_B+2\angle I_{B}BD.$$Because $\angle BD{I}_B$ and $\angle I_{B}BD$ are half of $\angle BDA$ and $\angle ABD$, respectively, the above expression can be simplified to$$\angle B{O}_{1}D=\angle B{O}_1{I}_B+\angle I_B{O}_{1}D=2\angle BD{I}_B+2\angle I_{B}BD=\angle ABD+\angle BDA.$$Similarly, $\angle C{O}_{2}D=\angle ACD+\angle CDA$. As a result$$\begin{array}{rl}\angle CPB& =\angle CPD+\angle BPD\\ & =\frac{1}{2}\cdot \angle C{O}_{2}D+\frac{1}{2}\cdot \angle B{O}_{1}D\\ & =\frac{1}{2}(\angle ABD+\angle BDA+\angle ACD+\angle CDA)\\ & =\frac{1}{2}(2\cdot 180^{\circ }-\angle BAC)\\ & =\frac{1}{2}\cdot 300^{\circ }=150^{\circ }.\end{array}$$ Therefore $\angle CPB$ is constant ($150^{\circ }$). Also, $P$ is $B$ or $C$ when $D$ is $B$ or $C$. Let point $L$ be on the same side of $\overline{BC}$ as $A$ with $LC=LB=BC=14$; $P$ is on the circle with $L$ as the center and $\overline{LC}$ as the radius, which is $14$. The shortest distance from $L$ to $\overline{BC}$ is $7\sqrt{3}$. When the area of $△BPC$ is the maximum, the distance from $P$ to $\overline{BC}$ has to be the greatest. In this case, it's $14-7\sqrt{3}$. The maximum area of $△BPC$ is$$\frac{1}{2}\cdot 14\cdot (14-7\sqrt{3})=98-49\sqrt{3}$$and the requested answer is $98+49+3=\boxed{150}$.