jmc

We first observe that by the Pythagorean theorem $△PBC$ must be a right triangle with right angle at $P$, since $PB=3$, $PC=4$, and $BC=5$.

$[△PBC]=\frac{1}{2}(3)(4)=6=\frac{1}{2}(PH)(5)$. Hence, the altitude $\overline{PH}$ from $P$ to $\overline{BC}$ has length $\frac{12}{5}$. Let $h$ be the length of the altitude from $A$ to $\overline{BC}$. Then $[△ABC]=\frac{1}{2}(h)(5)$, so the area is maximized when $A$ is most high above $\overline{BC}$. Since $AP=2$, maximization occurs when $A$ is directly over $P$, leading to a height of $h=\frac{12}{5}+2=\frac{22}{5}$. In this case, $$[△ABC]=\frac{1}{2}\left(\frac{22}{5}\right)(5)=\boxed{11}.$$