9. For the parabola $y^{2}=2 p x(p>0)$, a line $l$ passing through its focus $F$ intersects the parabola at points $A$ and $B$. Then the minimum value of $|F A| \cdot |F B|$ is ( ).
(A) $\frac{p^{2}}{4}$
(B) $p^{2}$
(C) $4 p^{2}$
(D) does not exist
9. B.

Focus $F\left(\frac{p}{2}, 0\right)$. When $l$ is not perpendicular to the $x$-axis, we can set the equation of the line $l$ as $y=k\left(x-\frac{p}{2}\right)$.
Substituting into the parabola equation and eliminating $y$, we get
$$
k^{2} x^{2}-p\left(k^{2}+2\right) x+\frac{k^{2} p^{2}}{4}=0 \text {. }
$$

The two real roots $x_{1}$ and $x_{2}$ are the $x$-coordinates of points $A$ and $B$.
By the relationship between roots and coefficients, we have
$$
\begin{array}{l}
x_{1}+x_{2}=\frac{p\left(k^{2}+2\right)}{k^{2}}=p+\frac{2 p}{k^{2}}, \\
x_{1} x_{2}=\frac{p^{2}}{4} .
\end{array}
$$

Also, $|F A|=x_{1}+\frac{p}{2},|F B|=x_{2}+\frac{p}{2}$, so
$$
\begin{array}{l}
|F A| \cdot|F B|=\left(x_{1}+\frac{p}{2}\right)\left(x_{2}+\frac{p}{2}\right) \\
=x_{1} x_{2}+\frac{p}{2}\left(x_{1}+x_{2}\right)+\frac{p^{2}}{4} \\
=\frac{p^{2}}{4}+\frac{p}{2}\left(p+\frac{2 p}{k^{2}}\right)+\frac{p^{2}}{4}=p^{2}+\frac{p^{2}}{k^{2}}>p^{2} .
\end{array}
$$

When $l$ is perpendicular to the $x$-axis, the $x$-coordinates of points $A$ and $B$ are both $\frac{p}{2}$, so,
$$
|F A| \cdot|F B|=\left(\frac{p}{2}+\frac{p}{2}\right)\left(\frac{p}{2}+\frac{p}{2}\right)=p^{2} .
$$

In summary, the minimum value of $|F A| \cdot|F B|$ is $p^{2}$.