15. As shown in the figure, $P$ 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 $\triangle P B C$.
Given $P\left(2 t^{2}, 2 t\right), M(1,0)$, the line through $P$ is $y-2 t=k\left(x-2 t^{2}\right)$ and it is tangent to the circle $M$, then we have
$$
\begin{array}{l}
\frac{\left|k\left(1-2 t^{2}\right)+2 t\right|}{\sqrt{1+k^{2}}}=1 \\
\Rightarrow 4 t^{2}\left(t^{2}-1\right) k^{2}-4 t\left(2 t^{2}-1\right) k+4 t^{2}-1=0
\end{array}
$$

Let the slopes of the lines $P B, P C$ be $k_{1}, k_{2}$, then
$$
\begin{array}{l}
y_{B}=2 t-2 t^{2} k_{1}, y_{C}=2 t-2 t^{2} k_{2}, \\
S_{\triangle P B C}=\frac{1}{2} \cdot 2 t^{2}\left|y_{B}-y_{C}\right|=2 t^{4}\left|k_{1}-k_{2}\right| \\
=2 t^{4} \cdot \frac{\sqrt{16 t^{2}\left(2 t^{2}-1\right)^{2}-16 t^{2}\left(t^{2}-1\right)\left(4 t^{2}-1\right)}}{4 t^{2}\left(t^{2}-1\right)} \\
=2 t^{2} \cdot \frac{|t| \sqrt{\left(2 t^{2}-1\right)^{2}-\left(t^{2}-1\right)\left(4 t^{2}-1\right)}}{t^{2}-1} \\
=\frac{2 t^{4}}{t^{2}-1}=2\left(\frac{t^{4}-1+1}{t^{2}-1}\right)=2\left(t^{2}+1+\frac{1}{t^{2}-1}\right) \\
=2\left(t^{2}-1+\frac{1}{t^{2}-1}+2\right) \geqslant 2(2+2)=8,
\end{array}
$$

Equality holds when $t^{2}=2 \Rightarrow t= \pm \sqrt{2}$. Therefore, the minimum area of $\triangle P B C$ is 8.