10. Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)
$$

respectively. If there exists a point $P$ on the right branch of the hyperbola such that
$$
\left(\overrightarrow{O P}+\overrightarrow{O F_{2}}\right) \cdot \overrightarrow{F_{2} P}=0 \text {, }
$$

where $O$ is the origin, and $\left|\overrightarrow{P F_{1}}\right|=\sqrt{3}\left|\overrightarrow{P F_{2}}\right|$, then the eccentricity of the hyperbola is ( ).
(A) $\sqrt{6}+\sqrt{2}$
(B) $\sqrt{3}+1$
(C) $\frac{\sqrt{3}+1}{2}$
(D) $\frac{\sqrt{6}+\sqrt{2}}{2}$
10. B.

From $\left(\overrightarrow{O P}+\overrightarrow{O F_{2}}\right) \cdot \overrightarrow{F_{2} P}=0$, we know
$$
\begin{array}{l}
\left(\overrightarrow{O P}+\overrightarrow{O F_{2}}\right) \cdot\left(\overrightarrow{O P}-\overrightarrow{O F_{2}}\right)=0 \\
\Rightarrow(\overrightarrow{O P})^{2}-\left(\overrightarrow{O F_{2}}\right)^{2}=0
\end{array}
$$

Thus, $|\overrightarrow{O P}|=\left|\overrightarrow{O F_{2}}\right|=\left|\overrightarrow{O F_{1}}\right|$.
Therefore, $\triangle P F_{1} F_{2}$ is a right triangle.
Also, $\left|\overrightarrow{P F_{1}}\right|=\sqrt{3}\left|\overrightarrow{P F_{2}}\right|$, then
$$
\angle F_{1}=30^{\circ}, \angle F_{2}=60^{\circ}, \angle F_{1} P F_{2}=90^{\circ} \text {. }
$$

By the definition of a hyperbola, we have
$$
\begin{array}{l}
\left|P F_{1}\right|-\left|P F_{2}\right|=2 a \\
\Rightarrow \sqrt{3} c-c=2 a \\
\Rightarrow \frac{c}{a}=\frac{2}{\sqrt{3}-1}=\sqrt{3}+1 .
\end{array}
$$