Answer: $12+\frac{69}{14} \sqrt{7}$
Solution:
$$
F B=\sqrt{A B^{2}-A F^{2}}=\sqrt{64-36}=2 \sqrt{7}
$$

The radius $O C$ is the midsegment of trapezoid $A D E B$. Hence $[A B E D]=4 \times 6=24$. Moreover
$$
\begin{aligned}
{[A B E D] } & =[A B F]+[A D E F] \\
24 & =\frac{2 \sqrt{7} \times 6}{2}+A D \times 6 \\
A D & =4-\sqrt{7} \\
E B & =A D+F B=4+\sqrt{7}
\end{aligned}
$$

Also by similarity (triangle $P A D$ and triangle $P B E$ )
$$
\begin{aligned}
\frac{P D}{A D} & =\frac{P E}{E B} \\
\frac{P D}{4-\sqrt{7}} & =\frac{P D+6}{4+\sqrt{7}} \\
4 P D+\sqrt{7} P D & =4 P D-\sqrt{7} P D+24-6 \sqrt{7} \\
2 \sqrt{7} P D & =24-6 \sqrt{7} \\
P D & =\frac{12}{\sqrt{7}}-3=\frac{12 \sqrt{7}}{7}-3
\end{aligned}
$$

Hence the area is equal to
$$
\frac{1}{2}\left(\frac{12 \sqrt{7}}{7}-3\right)(4+\sqrt{7})=12+\frac{69}{14} \sqrt{7}
$$
6. (Difficult) On the line containing diameter $\mathrm{AB}$ of a circle, a point $\mathrm{P}$ is chosen outside of this circle, with $\mathrm{P}$ closer to $\mathrm{A}$ than B. One of the two tangent lines through $\mathrm{P}$ is drawn. Let $D$ and $E$ be two points on the tangent line such that $A D$ and $B E$ are perpendicular to it. If $D E=6$, find the area of triangle $B E P$.