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

$O$ is the origin, point $P$ is on the left branch of the hyperbola, point $M$ is on the right directrix, and it satisfies
$$
F_{1} \boldsymbol{O}=\boldsymbol{P M}, O P=\lambda\left(\frac{O F_{1}}{\left|O F_{1}\right|}+\frac{O M}{|O M|}\right)(\lambda>0) \text {. }
$$
(1) Find the eccentricity of this hyperbola;
(2) If this hyperbola passes through the point $N(2, \sqrt{3}), B_{1}(0, b)$, and $B_{2}(0, -b)$, and points $A$ and $B$ are on the hyperbola, and $B_{2} \boldsymbol{A}=\mu \boldsymbol{B}_{2} B(\mu \in \mathbf{R})$, when $B_{1} A \cdot B_{1} B=0$, find the equation of the line $A B$.
16. (1) From $\boldsymbol{F}_{1} \boldsymbol{O}=\boldsymbol{P M}$, we know that quadrilateral $P F_{1} O M$ is a parallelogram.

Also, $\boldsymbol{O P}=\lambda\left(\frac{O F_{1}}{\left|O F_{1}\right|}+\frac{\boldsymbol{O M}}{|O M|}\right)(\lambda>0)$, so $O P$ is the angle bisector of $\angle F_{1} O M$.
Therefore, quadrilateral $P F_{1} O M$ is a rhombus.
From $\left|O F_{1}\right|=c,\left|P F_{1}\right|=|P M|=c$, we get
$\left|\boldsymbol{P F}_{2}\right|=2 a+c$.
Also, $\frac{\left|\boldsymbol{P} \boldsymbol{F}_{2}\right|}{|\boldsymbol{P M}|}=e$, then $\frac{2 a+c}{c}=e$, which simplifies to $1+\frac{2}{e}=e$.
Solving, we get $e=2, e=-1$ (discard).
Therefore, $e=2$.
(2) From $e=2$, let $c=2 m, a=m(m>0)$. Then $b=\sqrt{3} m$.
The hyperbola equation is $\frac{x^{2}}{m^{2}}-\frac{y^{2}}{3 m^{2}}=1$.
Substituting $N(2, \sqrt{3})$ into the equation, we get $m^{2}=3$.
Thus, the hyperbola equation is $\frac{x^{2}}{3}-\frac{y^{2}}{9}=1$.
Therefore, $B_{1}(0,3) 、 B_{2}(0,-3)$.
From $\boldsymbol{B}_{2} \boldsymbol{A}=\mu \boldsymbol{\boldsymbol { B } _ { 2 } \boldsymbol { B }}$, we know that $A 、 B_{2} 、 B$ are collinear.
Let the equation of line $A B$ be $y=k x-3$, and let $A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right)$.
Substituting $y=k x-3$ into $\frac{x^{2}}{3}-\frac{y^{2}}{9}=1$, we get $\left(3-k^{2}\right) x^{2}+6 k x-18=0$.
By Vieta's formulas, we have
$x_{1}+x_{2}=\frac{6 k}{k^{2}-3}, x_{1} x_{2}=\frac{18}{k^{2}-3}$.
From $\boldsymbol{B}_{1} \boldsymbol{A} \cdot \boldsymbol{B}_{1} \boldsymbol{B}=0$, we have
$$
x_{1} x_{2}+\left(k x_{1}-6\right)\left(k x_{2}-6\right)=0 \text {. }
$$

Rearranging, we get
$$
\left(1+k^{2}\right) x_{1} x_{2}-6 k\left(x_{1}+x_{2}\right)+36=0 \text {. }
$$

Substituting $x_{1}+x_{2}, x_{1} x_{2}$, we get
$$
\frac{18\left(1+k^{2}\right)}{k^{2}-3}-\frac{36 k^{2}}{k^{2}-3}+36=0 \text {. }
$$

Solving, we get $k= \pm \sqrt{5}$.
Therefore, the equation of the line is $y= \pm \sqrt{5} x-3$.