Example 6 As shown in Figure 6, it is known that $\triangle A B C$ is inscribed in $\odot O$. A line $l$ is drawn through the midpoint $D$ of $B C$, parallel to $A C$. Line $l$ intersects $A B$ at point $E$, and intersects $\odot O$ at points $G$ and $F$, and intersects the tangent line of $\odot O$ at point $A$ at point $P$. If $P E=3, E D=2, E F$ $=3$, then the length of $P A$ is ( ).
(A) $\sqrt{5}$
(B) $\sqrt{6}$
(C) $\sqrt{7}$
(D) $2 \sqrt{2}$
(2011, Joint Autonomous Admission Examination of Excellent Talent Training Cooperative Universities).
Since $A C / / P F$, we have,
$$
\begin{array}{l}
\angle B D E=\angle B C A=\angle P A E \\
\Rightarrow P, A, D, E \text{ are concyclic} \\
\Rightarrow B E \cdot A E=P E \cdot D E=6 \\
\Rightarrow B E=A E=\sqrt{6} .
\end{array}
$$

By the intersecting chords theorem,
$$
G E \cdot E F=B E^{2} \text{. }
$$

Thus, $G E=2, P G=1$.
By the secant-tangent theorem, $P A^{2}=P G \cdot P F$.
Therefore, $P A=\sqrt{6}$.