8. As shown in Figure 1, $P$ is a point inside $\triangle ABC$, with the perimeter and area of $\triangle ABC$ being $l$ and $s$ respectively. The perpendicular segments from point $P$ to $AB$, $BC$, and $CA$ are $PD$, $PE$, and $PF$ respectively, and
$$
\frac{AB}{PD}+\frac{BC}{PE}+\frac{CA}{PF} \leqslant \frac{l^{2}}{2s} \text{. }
$$

Then $P$ is the 
center among the "five
centers" of $\triangle ABC$.
8. Inner.

From the area formula we get
$$
A B \cdot P D+B C \cdot P E+A C \cdot P F=2 s \text {. }
$$

Using the Cauchy-Schwarz inequality we get
$$
\begin{array}{l}
(A B \cdot P D+B C \cdot P E+A C \cdot P F) \cdot \\
\left(\frac{A B}{P D}+\frac{B C}{P E}+\frac{C A}{P F}\right) \\
\geqslant(A B+B C+C A)^{2}=l^{2} .
\end{array}
$$

From the given condition, we know that the equality holds in the above inequality.
Thus, $P D=P E=P F$.
Therefore, $P$ is the incenter of $\triangle A B C$.