15. As shown in Figure $1, P$ 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 $\triangle P B C$.
15. Let \( P\left(x_{0}, y_{0}\right) \), \( B(0, b) \), and \( C(0, c) \), and assume \( b > c \).
The equation of line \( PB \) is \( y - b = \frac{y_{0} - b}{x_{0}} x \).
Simplifying, we get \( \left(y_{0} - b\right) x - x_{0} y + x_{0} b = 0 \).
The distance from the circle center \((1,0)\) to \( PB \) is 1, so
\[ \frac{\left|y_{0} - b + x_{0} b\right|}{\sqrt{\left(y_{0} - b\right)^{2} + x_{0}^{2}}} = 1. \]
Thus, \( \left(y_{0} - b\right)^{2} + x_{0}^{2} \)
\[ = \left(y_{0} - b\right)^{2} + 2 x_{0} b \left(y_{0} - b\right) + x_{0}^{2} b^{2}. \]
It is known that \( x_{0} > 2 \). Simplifying the above equation, we get
\[ \left(x_{0} - 2\right) b^{2} + 2 y_{0} b - x_{0} = 0. \]
Similarly, \( \left(x_{0} - 2\right) c^{2} + 2 y_{0} c - x_{0} = 0 \).
Therefore, \( b + c = \frac{-2 y_{0}}{x_{0} - 2} \), \( bc = \frac{-x_{0}}{x_{0} - 2} \).
Thus, \( (b - c)^{2} = \frac{4 x_{0}^{2} + 4 y_{0}^{2} - 8 x_{0}}{\left(x_{0} - 2\right)^{2}} \).
Since \( P\left(x_{0}, y_{0}\right) \) is a point on the parabola, we have \( y_{0}^{2} = 2 x_{0} \), so
\[ (b - c)^{2} = \frac{4 x_{0}^{2}}{\left(x_{0} - 2\right)^{2}}, \quad b - c = \frac{2 x_{0}}{x_{0} - 2}. \]
Therefore, \( S_{\triangle P B C} = \frac{1}{2} (b - c) x_{0} = \frac{x_{0}}{x_{0} - 2} \cdot x_{0} \)
\[ = \left(x_{0} - 2\right) + \frac{4}{x_{0} - 2} + 4 \geqslant 2 \sqrt{4} + 4 = 8. \]
When \( \left(x_{0} - 2\right)^{2} = 4 \), the above inequality holds with equality, at which point,
\[ x_{0} = 4, \quad y_{0} = \pm 2 \sqrt{2}. \]
Therefore, the minimum value of \( S_{\triangle P B C} \) is 8.