smc

Given leg length $x$, we can write the perimeter of this triangle to be $2x+x\sqrt{2}=2p$. Thus, $x(2+\sqrt{2})=2p$. Divide to get $x=\frac{2p}{2+\sqrt{2}}$. Multiply by the conjugate and simplify to get $x=p(2-\sqrt{2})$. Square and divide by two to get the area of the triangle, or $\frac{p^2\ast (2-{\sqrt{2}}^2)}{2}$, or $p^2(3-2\sqrt{2})$. $\boxed{(3-2\sqrt{2})p^2}$