4. As shown in Figure 1, in the tetrahedron $P-ABC$, the three lateral edges $PA$, $PB$, and $PC$ are mutually perpendicular. The dihedral angles formed by the lateral faces $PAB$, $PBC$, and $PCA$ with the base $ABC$ have plane angles of $\theta_{1}$, $\theta_{2}$, and $\theta_{3}$, respectively. The area of the base $\triangle ABC$ is $4 \sqrt{3}$. If
$$
\tan \theta_{1}+\tan \theta_{2}+\tan \theta_{0}=3 \sqrt{2},
$$
then the correct statement about $V_{P-ABC}$ is ( ).
(A) equals $2 \sqrt{3}$
(B) equals $\sqrt{6}$
(C) equals $\frac{8 \sqrt{2}}{3}$
(D) the conditions are insufficient, $V_{P-ABC}$ cannot be determined
4. C.

As shown in Figure 5, let's assume
$$
\begin{array}{l}
P A=a, P B=b, \\
P C=c. \text{ Draw } P N \perp \\
A B \text{ at } N \text{, and connect } C N. \\
\text{ It is easy to see that } \angle P N C=\theta_{1} \\
\Rightarrow \tan \theta_{1} \\
=\tan \angle P N C \\
=\frac{P C}{P N} \\
=\frac{P C}{\frac{2 S_{\triangle P A B}}{A B}}=\frac{c}{\frac{a b}{\sqrt{a^{2}+b^{2}}}}=\frac{c}{a b} \sqrt{a^{2}+b^{2}}. \\
\end{array}
$$

Similarly, $\tan \theta_{2}=\frac{a}{b c} \sqrt{b^{2}+c^{2}}$,
$$
\tan \theta_{3}=\frac{b}{c a} \sqrt{c^{2}+a^{2}}.
$$

Therefore, $\tan \theta_{1} \cdot \tan \theta_{2} \cdot \tan \theta_{3}$
$$
\begin{array}{l}
=\frac{1}{a b c} \sqrt{a^{2}+b^{2}} \cdot \sqrt{b^{2}+c^{2}} \cdot \sqrt{c^{2}+a^{2}} \\
\geqslant \frac{1}{a b c} \sqrt{2 a b} \cdot \sqrt{2 b c} \cdot \sqrt{2 c a}=2 \sqrt{2}.
\end{array}
$$

The equality holds if and only if $a=b=c$.
Also, $3 \sqrt{2}=\tan \theta_{1}+\tan \theta_{2}+\tan \theta_{3}$
$$
\geqslant 3 \sqrt[3]{\tan \theta_{1} \cdot \tan \theta_{2} \cdot \tan \theta_{3}} \geqslant 3 \sqrt{2},
$$

The equality holds if and only if $\tan \theta_{1}=\tan \theta_{2}=\tan \theta_{3}$, i.e., $a=b=c$.
From $a=b=c$, we get $A B=B C=C A$.
Given $S_{\triangle A B C}=4 \sqrt{3}$, we have
$$
\frac{\sqrt{3}}{4} A B^{2}=4 \sqrt{3} \Rightarrow A B=4.
$$

Thus, $a=b=c=2 \sqrt{2}$.
Therefore, $V_{P-A B C}=V_{C-P A B}=\frac{1}{6} a b c=\frac{8 \sqrt{2}}{3}$.