9. In the quadrilateral pyramid $P-ABCD$,
$$
\begin{array}{l}
\overrightarrow{AB}=(4,-2,3), \overrightarrow{AD}=(-4,1,0), \\
\overrightarrow{AP}=(-6,2,-8) .
\end{array}
$$

Then the height $h$ of this quadrilateral pyramid is $(\quad)$.
(A) 1
(B) 2
(C) 13
(D) 26
9. B.

Let the height of the quadrilateral pyramid be $P H, \overrightarrow{P H}=(x, y, z)$. Then $\overrightarrow{P H} \cdot \overrightarrow{A B}=4 x-2 y+3 z=0$,
$\overrightarrow{P H} \cdot \overrightarrow{A D}=-4 x+y=0$.
Solving these, we get $y=4 x, z=\frac{4}{3} x$.
$\overrightarrow{P H}=\left(x, 4 x, \frac{4}{3} x\right)$.
At this point, $(\overrightarrow{A P}+\overrightarrow{P H}) \cdot \overrightarrow{P H}=\overrightarrow{A P} \cdot \overrightarrow{P H}+|\overrightarrow{P H}|^{2}=0$.
$\overrightarrow{A P} \cdot \overrightarrow{P H}=-6 x+8 x-\frac{32}{3} x=-\frac{26}{3} x$,
$|\overrightarrow{P H}|^{2}=x^{2}+16 x^{2}+\frac{16}{9} x^{2}=\frac{169}{9} x^{2}$.
Therefore, $-\frac{26}{3} x+\frac{169}{9} x^{2}=0$.
Solving for $x$, we get $x=0$ (discard), $x=\frac{6}{13}$.
Thus, $h=|\overrightarrow{P H}|=\sqrt{x^{2}+16 x^{2}+\frac{16}{9} x^{2}}$
$$
=|x| \cdot \sqrt{\frac{169}{9}}=\frac{6}{13} \times \frac{13}{3}=2 \text {. }
$$