Answer: 112

Small Hint:

Write each triangle’s area as one-half the product of its two side segments and $\sin\theta,$ where $\theta$ is an angle at $P;$ the area condition factors as $(AP - CP)(DP - BP) = 0$

Big Hint:

Take $AP = CP$ and apply the law of cosines to all four triangles at $P;$ add and subtract in pairs, using that the angles at $P$ are supplementary

Solution:

Let $a = AP,$ $b = BP,$ $c = CP,$ $d = DP,$ and let $\theta = \angle CPD.$ Since $\sin(\pi - \theta) = \sin\theta,$ the equal-area condition $\frac{1}{2}(ab + cd)\sin\theta$ $= \frac{1}{2}(ad + bc)\sin\theta$ simplifies to $(a - c)(d - b) = 0.$ By symmetry assume $a = c.$

The law of cosines in triangles $BPC$ and $APB$ (whose angles at $P$ are supplementary) gives $a^2 + b^2 + 2ab\cos\theta = 196$ and $a^2 + b^2 - 2ab\cos\theta = 100,$ so $a^2 + b^2 = 148$ and $ab\cos\theta = 24.$ Similarly triangles $APD$ and $CPD$ give $a^2 + d^2 = 180$ and $ad\cos\theta = 40.$ Dividing, $\frac{d}{b} = \frac{5}{3},$ while subtracting gives $d^2 - b^2 = 32;$ hence $b = 3\sqrt{2},$ $d = 5\sqrt{2},$ $a^2 = 130,$ and $\cos^2\theta = \frac{24^2}{130 \cdot 18} = \frac{16}{65},$ so $\sin\theta = \frac{7}{\sqrt{65}}.$

The total area is $$\begin{aligned} \frac{1}{2}(a + c)(b + d) \\ &\quad {}\cdot \sin\theta \\ &= a(b + d)\sin\theta \\ &= \sqrt{130} \cdot 8\sqrt{2} \\ &\quad {}\cdot \frac{7}{\sqrt{65}} \\ &= 112. \end{aligned}$$