16. As shown in Figure $2, P$ is a moving point on the parabola $y^{2}=2 x$, points $B$ and $C$ are on the $y$-axis, and the circle $(x-1)^{2}+y^{2}=1$ is inscribed in $\triangle P B C$. Find the minimum value of the area of $\triangle P B C$.
16. Let \( P\left(x_{0}, y_{0}\right) \), \( B(0, b) \), \( C(0, c) \), and assume \( b > c \). The line \( l_{P B} \) is given by \( y - b = \frac{y_{0} - b}{x_{0}} x \), which can be rewritten as
\[
\left(y_{0} - b\right) x - x_{0} y + x_{0} b = 0.
\]

The distance from the circle's center \((1, 0)\) to \( P B \) is 1, so
\[
\frac{\left|y_{0} - b + x_{0} b\right|}{\sqrt{\left(y_{0} - b\right)^{2} + x_{0}^{2}}} = 1.
\]

Thus,
\[
\left(y_{0} - b\right)^{2} + x_{0}^{2} = \left(y_{0} - b\right)^{2} + 2 x_{0} b \left(y_{0} - b\right) + x_{0}^{2} b^{2}.
\]

Given \( x_{0} > 2 \), the equation simplifies to
\[
\left(x_{0} - 2\right) b^{2} + 2 y_{0} b - x_{0} = 0.
\]

Similarly,
\[
\left(x_{0} - 2\right) c^{2} + 2 y_{0} c - x_{0} = 0.
\]

Therefore,
\[
b + c = \frac{-2 y_{0}}{x_{0} - 2}, \quad b c = \frac{-x_{0}}{x_{0} - 2}.
\]

Thus,
\[
(b - c)^{2} = \frac{4 x_{0}^{2} + 4 y_{0}^{2} - 8 x_{0}}{\left(x_{0} - 2\right)^{2}}.
\]

Since \( P\left(x_{0}, y_{0}\right) \) lies on the parabola, \( y_{0}^{2} = 2 x_{0} \). Therefore,
\[
(b - c)^{2} = \frac{4 x_{0}^{2}}{\left(x_{0} - 2\right)^{2}} \Rightarrow b - c = \frac{2 x_{0}}{x_{0} - 2}.
\]

Hence,
\[
S_{\triangle P B C} = \frac{1}{2} (b - c) x_{0} = \frac{x_{0}}{x_{0} - 2} \cdot x_{0} = \left(x_{0} - 2\right) + \frac{4}{x_{0} - 2} + 4 \geqslant 2 \sqrt{4} + 4 = 8.
\]

The equality holds when \( \left(x_{0} - 2\right)^{2} = 4 \), which gives \( x_{0} = 4 \) and \( y_{0} = \pm 2 \sqrt{2} \).

Therefore, the minimum value of \( S_{\triangle P B C} \) is 8.
(Provided by Shi Dongyang)