6. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. If $\angle B=60^{\circ}, O$ is the circumcenter of $\triangle A B C$, point $P$ is on the plane of $\triangle A B C$, $O P=O A + O B + O C$, and $B P \cdot B C=8$. Then the maximum value of the altitude $h$ on side $A C$ is ( ).
(A) $2 \sqrt{3}$
(B) $\sqrt{3}$
(C) $2 \sqrt{2}$
(D) $\sqrt{2}$
6. A.

From the given, $O P-O A=O B+O C$, which means
$$
\begin{array}{l}
A P=O B+O C . \\
\text { Also, }(O B+O C) \cdot B C \\
=(O C+O B) \cdot(O C-O B) \\
=O C^{2}-O B^{2}=|O C|^{2}-|O B|^{2}=0,
\end{array}
$$

Thus, $(O B+O C) \perp B C$.
From equation (1), we know $\boldsymbol{A P} \perp B C$.
Similarly, $B P \perp A C$.
Therefore, $P$ is the orthocenter of $\triangle A B C$.
Given $B P \cdot B C=8$, which means
$$
\begin{array}{l}
|B P| \cdot|B C| \cos \angle P B C \\
=(|B P| \cos \angle P B C) a \\
=(|A B| \cos B)_{a}=a c \cos 60^{\circ}=8,
\end{array}
$$

Thus, $a c=16$.
Hence, $S_{\triangle A B C}=\frac{1}{2} a c \sin B=4 \sqrt{3}=\frac{1}{2} b h$.
Clearly, $h=\frac{8 \sqrt{3}}{b}$.
By the cosine rule,
$$
\begin{array}{l}
b^{2}=a^{2}+c^{2}-2 a c \cos B \\
=a^{2}+c^{2}-16 \geqslant 2 a c-16,
\end{array}
$$

Thus, $b^{2} \geqslant 2 a c-16=16 \Rightarrow b \geqslant 4$.
When and only when $a=c$, $b$ achieves its minimum value of 4. From equation (2), the maximum value of $h$ is $2 \sqrt{3}$.