8. Given that $P$ is a moving point on the incircle of the equilateral $\triangle A B C$ with side length $6 \sqrt{3}$. Then the minimum value of $B P+\frac{1}{2} P C$ is $\qquad$
8. $\frac{3 \sqrt{21}}{2}$.

As shown in Figure 5, let the incircle of $\triangle ABC$ be $\odot O$, and take point $M$ on $OC$ such that $OM = \frac{1}{4} OC$.
Given that the side length of the equilateral $\triangle ABC$ is $6 \sqrt{3}$, we know
$$
\begin{array}{l}
OC = 6, OP = 3, OM = \frac{3}{2} \\
\Rightarrow \frac{OM}{OP} = \frac{OP}{OC} = \frac{1}{2} \Rightarrow \triangle OPM \sim \triangle OCP \\
\Rightarrow \frac{PM}{PC} = \frac{OM}{OP} = \frac{1}{2} \Rightarrow PM = \frac{1}{2} PC.
\end{array}
$$

Notice that, $PB + \frac{1}{2} PC = PB + PM \geqslant BM$.
Draw $MH \perp BC$ at point $H$. Then
$$
\begin{array}{l}
MH = \frac{1}{2} MC = \frac{1}{2}(OC - OM) = \frac{9}{4}, \\
BH = BC - CH = 6 \sqrt{3} - \frac{9 \sqrt{3}}{4} = \frac{15 \sqrt{3}}{4}. \\
\text{Thus, } BM = \sqrt{\left(\frac{15 \sqrt{3}}{4}\right)^2 + \left(\frac{9}{4}\right)^2} \\
= \frac{\sqrt{756}}{4} = \frac{3 \sqrt{21}}{2}.
\end{array}
$$

Therefore, $BP + \frac{1}{2} PC = PB + PM \geqslant BM = \frac{3 \sqrt{21}}{2}$.
When the moving point $P$ is the intersection of $BM$ and the incircle, the minimum value $\frac{3 \sqrt{21}}{2}$ is obtained.