jmc

We first extend line segment $\overline{CP}$ so that it intersects $\overline{AB}$. We'll call this point of intersection point $E$, so $\overline{CE}$ is a perpendicular bisector to segment $\overline{AB}$ and $AE=EB=3$. We also let $x=$ the lengths of segments $\overline{PA}$, $\overline{PB}$, and $\overline{PC}$, so line segment $\overline{PE}$ will have length $6-x$. Now we have that $△AEP$ is a right triangle. Using the Pythagorean Theorem and solving for $x$, we have that: $$\begin{array}{rl}& A{E}^2+P{E}^2=P{A}^2\\ \Rightarrow & 3^2+(6-x{)}^2=x^2\\ \Rightarrow & 9+36-12x+x^2=x^2\\ \Rightarrow & 12x=45\\ \Rightarrow & x=\frac{15}{4}.\end{array}$$ Thus, $△APB$ has base $6$ and a a height of $6-x=6-\frac{15}{4}=\frac{9}{4}$. It follows that $△APB$ has an area of $\frac{1}{2}bh=\frac{1}{2}\cdot 6\cdot \left(\frac{9}{4}\right)=\boxed{\frac{27}{4}}$ square inches.