Solution:

In triangle $ABD$ the incenter satisfies $\angle B I_B D = 90^\circ + \frac{\angle BAD}{2},$ and likewise $\angle C I_C D = 90^\circ + \frac{\angle DAC}{2},$ so these two angles sum to $180^\circ + \frac{\angle BAC}{2}.$ The law of cosines gives $\cos \angle BAC = \frac{10^2 + 16^2 - 14^2}{2 \cdot 10 \cdot 16} = \frac{1}{2},$ so $\angle BAC = 60^\circ$ and the sum is $210^\circ.$

The second intersection point $P$ lies on the opposite side of $\overline{BC}$ from the incenters (were it on the same side, the two cyclic quadrilaterals would force $\angle BPC = 210^\circ \gt 180^\circ$). Then $BI_BDP$ and $CI_CDP$ are convex cyclic quadrilaterals, so $$\angle BPC = \angle BPD + \angle DPC = \left(180^\circ - \angle BI_BD\right) + \left(180^\circ - \angle CI_CD\right) = 360^\circ - 210^\circ = 150^\circ,$$ independent of $D.$ Hence $P$ moves along a fixed circular arc through $B$ and $C.$

The area of triangle $BPC$ is maximized at the midpoint of the arc, where $BP = PC = x.$ The law of cosines gives $14^2 = 2x^2 + \sqrt{3}\,x^2,$ so $x^2 = \frac{196}{2 + \sqrt{3}} = 196\left(2 - \sqrt{3}\right),$ and the area is $\frac{1}{2}x^2 \sin 150^\circ = 49\left(2 - \sqrt{3}\right) = 98 - 49\sqrt{3}.$ Thus $a + b + c = 98 + 49 + 3 = 150.$