8. 8 .

Let point $P\left(2 t^{2}, 2 t\right)$.
Since it is an incircle, we know $|t|>1$.
Let the slope of the line passing through point $P$ be $k$. Then the equation of the line is
$$
y-2 t=k\left(x-2 t^{2}\right) \text {. }
$$

Since this line is tangent to the circle, we have
$$
d=\frac{\left|2 t^{2} k-2 t-k\right|}{\sqrt{1+k^{2}}}=1,
$$

which simplifies to $\left(4 t^{4}-4 t^{2}\right) k^{2}-4\left(2 t^{2}-1\right) t k+4 t^{2}-1=0$.
By Vieta's formulas, we can express $k_{1}+k_{2}$ and $k_{1} k_{2}$ in terms of $\iota$, and calculate
$$
\begin{array}{l}
\left|y_{B}-y_{C}\right|=\frac{2 t^{2}}{t^{2}-1} . \\
\text { Hence } S_{\triangle P B C}=2 \frac{t^{4}}{t^{2}-1}=2\left(t^{2}-1+\frac{1}{t^{2}-1}+2\right) \\
\geqslant 8,
\end{array}
$$

and when $t^{2}=2$, $S_{\triangle P B C}$ reaches its minimum value of 8.
8. Given that $P$ is a moving point on the parabola $y^{2}=2 x$, points $B$ and $C$ are on the $y$-axis, and $(x-1)^{2}+y^{2}=1$ is the incircle of $\triangle P B C$. Then the minimum value of $S_{\triangle P B C}$ is $\qquad$