9 In the quadrilateral pyramid $P-ABCD$, $\overrightarrow{AB}=(4,-2,3)$, $\overrightarrow{AD}=(-4,1,0)$, $\overrightarrow{AP}=(-6,2,-8)$, then the height $h$ of this quadrilateral pyramid is $h=(\quad)$.
(A) 1
(B) 2
(C) 13
(D) 26
(9) Let the height of the pyramid be $P H, \overrightarrow{P H}=(x, y, z)$, then
$$
\begin{array}{l}
\overrightarrow{P H} \cdot \overrightarrow{A B}=4 x-2 y+3 z=0, \\
\overrightarrow{P H} \cdot \overrightarrow{A D}=-4 x+y=0,
\end{array}
$$

Solving, we get $y=4 x, z=\frac{4}{3} x$. Therefore,
$$
\overrightarrow{P H}=\left(x, 4 x, \frac{4}{3} x\right)
$$

At this point, we should have
$$
(\overrightarrow{A P}-\overrightarrow{P H}) \cdot \overrightarrow{P H}=\overrightarrow{A P} \cdot \overrightarrow{P H}-|\overrightarrow{P H}|^{2}=0
$$

Since
$$
\begin{array}{l}
\overrightarrow{A P} \cdot \overrightarrow{P H}=-6 x+8 x-\frac{32}{3} x=-\frac{26}{3} x, \\
|\overrightarrow{P H}|^{2}=x^{2}+16 x^{2}+\frac{16}{9} x^{2}=\frac{169}{9} x^{2},
\end{array}
$$

We have $-\frac{26}{3} x-\frac{169}{9} x^{2}=0$. Solving, we get $x=0$ (discard), $x=-\frac{6}{13}$. Therefore,
$$
\begin{aligned}
h & =|\overrightarrow{P H}| \\
& =\sqrt{x^{2}+16 x^{2}+\frac{16}{9} x^{2}} \\
& =|x| \cdot \sqrt{\frac{169}{9}} \\
& =\frac{6}{13} \cdot \frac{13}{3} \\
& =2 .
\end{aligned}
$$

Hence, the answer is B.