jmc

Notice that $\overline{BP}$ bisects the right angle at $B$. Thus, the Angle Bisector Theorem tells us that $AB/BC=AP/PC=1/2$. So, we have $AB=x$ and $BC=2x$ for some $x$. By the Pythagorean theorem, we have $5x^2=A{C}^2=9$, so $x^2=\frac{9}{5}$. Finally, the desired area $\frac{1}{2}(x)(2x)=x^2=\boxed{\frac{9}{5}}$.