(16) (This sub-question is worth 20 points)

Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with eccentricity $e=\frac{1}{2}$, the endpoints of the minor axis are $B_{1}, B_{2}$, and the foci are $F_{1}, F_{2}$. The radius of the incircle of the quadrilateral $F_{1} B_{1} F_{2} B_{2}$ is $\frac{\sqrt{3}}{2}$.
(I) Find the equation of the ellipse $C$;
(II) A line passing through the left focus $F_{1}$ intersects the ellipse at points $M, N$ and the line $x=-4$ at point $P$. Let $\overrightarrow{P M}=\lambda \overrightarrow{M F_{1}}, \overrightarrow{P N}=\mu \overrightarrow{N F_{1}}$. Prove that $\lambda+\mu$ is a constant, and find this constant.
Solution:
(I) As shown in the figure,

Let the point of tangency of the incircle of quadrilateral $F_{1} B_{1} F_{2} B_{2}$ with side $B_{1} B_{2}$ be $G$,
connect $O G,|O G|=\frac{\sqrt{3}}{2}$.
From $S_{\triangle O B_{2} F_{2}}=\frac{1}{2}\left|O B_{2}\right| \cdot\left|O F_{2}\right|=\frac{1}{2}\left|B_{2} F_{2}\right| \cdot|O G|,\left|O B_{2}\right|=b,\left|O F_{2}\right|=c,\left|B_{2} F_{2}\right|=a$,
we get $b c=\frac{\sqrt{3}}{2} a$.

Also, $e=\frac{c}{a}=\frac{1}{2}, a^{2}=b^{2}+c^{2}$, solving these gives $a=2, b=\sqrt{3}$, hence the equation of the ellipse $C$ is $C: \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.
(II) According to the given conditions, we can set the equation of line $M N$ as $y=k(x+1)$,

Substituting into the equation of the ellipse, we get $\left(3+4 k^{2}\right) x^{2}+8 k^{2} x+4\left(k^{2}-3\right)=0$.

Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$, then $x_{1}+x_{2}=-\frac{8 k^{2}}{3+4 k^{2}}, x_{1} x_{2}=\frac{4\left(k^{2}-3\right)}{3+4 k^{2}}$.
Also, $P(-4,-3 k), F_{1}(-1,0)$, from $\overline{P M}=\lambda \overrightarrow{M F_{1}}, \overrightarrow{P N}=\mu \overrightarrow{N F_{1}}$,
we get $\lambda=-\frac{x_{1}+4}{x_{1}+1}, \mu=-\frac{x_{2}+4}{x_{2}+1}$.
$\therefore \lambda+\mu=-\frac{x_{1}+4}{x_{1}+1}-\frac{x_{2}-4}{x_{2}+1}=-\frac{2 x_{1} x_{2}+5\left(x_{1}+x_{2}\right)+8}{\left(x_{1}+1\right)\left(x_{2}+1\right)}$,
$$
\begin{aligned}
\because 2 x_{1} x_{2}+5\left(x_{1}+x_{2}\right)+8= & 2 \cdot \frac{4\left(k^{2}-3\right)}{3+4 k^{2}}+5 \cdot\left(-\frac{8 k^{2}}{3+4 k^{2}}\right)+8 \\
& =\frac{8 k^{2}-24-40 k^{2}+24+32 k^{2}}{3+4 k^{2}}=0,
\end{aligned}
$$
$\therefore \lambda+\mu=0$ is a constant value.