Solution:

Let $\omega$ denote the common angle and $L = AB = AC.$ Since $\angle PAB = \omega,$ we have $\angle PAC = 90^\circ - \omega,$ and with $\angle PCA = \omega$ the angles of triangle $APC$ give $\angle APC = 90^\circ.$ Hence in right triangle $APC,$ $$L = AC = \frac{AP}{\sin\omega} = \frac{10}{\sin\omega}.$$

In triangle $ABP,$ the angle at $A$ is $\omega$ and the angle at $B$ is $45^\circ - \omega,$ so $\angle APB = 135^\circ.$ The law of sines gives $\frac{AP}{\sin(45^\circ - \omega)} = \frac{AB}{\sin 135^\circ},$ that is, $10 \sin 135^\circ = L \sin(45^\circ - \omega).$ Substituting $L = \frac{10}{\sin\omega}$ and expanding yields $$\sin\omega = \sqrt{2}\,\sin(45^\circ - \omega) = \cos\omega - \sin\omega,$$ so $\tan\omega = \frac{1}{2}$ and $\sin^2\omega = \frac{1}{5}.$

Therefore $L^2 = \frac{100}{\sin^2\omega} = 500,$ and the area is $\frac{1}{2}L^2 = 250.$