Inradius and Area of Right Isosceles Triangle

Since $P$ is equidistant from all three sides, $P$ is the incenter of $\triangle ABC$, and the given perpendicular distance is the inradius $r = 4(\sqrt{2} - 1)\text{ cm}$.

For a right-angled isosceles triangle with legs $a, a$ and hypotenuse $a\sqrt{2}$: $r = \frac{a + a - a\sqrt{2}}{2} = \frac{a(2 - \sqrt{2})}{2} = a\left(1 - \frac{\sqrt{2}}{2}\right) = \frac{a(\sqrt{2} - 1)}{\sqrt{2}}$

Equating to given $r$: $\frac{a(\sqrt{2} - 1)}{\sqrt{2}} = 4(\sqrt{2} - 1) \implies a = 4\sqrt{2}\text{ cm}$

Area of $\triangle ABC = \frac{1}{2} a^2 = \frac{1}{2} (4\sqrt{2})^2 = \frac{1}{2} (32) = 16\text{ sq. cm}$.