Example 5 As shown in Figure 4, if $P$ is any point inside the equilateral $\triangle A B C$, and perpendiculars are drawn from $P$ to the sides $A B$, $B C$, and $C A$, with the feet of the perpendiculars being $D$, $E$, and $F$ respectively, then $\frac{A D+B E+C F}{P D+P E+P F}=$ ( ).
(A) 2
(B) $\sqrt{3}$
(C) $2 \sqrt{3}$
(D) $2 \sqrt{2}$
Solution 1 As shown in Figure 4, let the side length of the equilateral $\triangle ABC$ be $a$. Draw $A_{1} B_{1} \parallel AB, B_{2} C_{2} \parallel BC, C_{1} A_{2} \parallel CA$, intersecting sides $BC$, $CA$, and $AB$ at points $B_{1}$ and $C_{1}$, $A_{1}$ and $C_{2}$, $A_{2}$ and $B_{2}$, respectively. It is easy to see that,
$$
\begin{array}{l}
A_{1} C_{2} + A_{2} P + P C_{1} = A_{1} C_{2} + A A_{1} + C C_{2} = a. \\
\text{Then } P D + P E + P F \\
= \frac{\sqrt{3}}{2} A_{2} P + \frac{\sqrt{3}}{2} P C_{1} + \frac{\sqrt{3}}{2} A_{1} C_{2} \\
= \frac{\sqrt{3}}{2} a, \\
A D + B E + C F \\
= A A_{2} + B B_{1} + C C_{1} + \frac{1}{2} \left( A_{2} P + B_{1} P + P C_{2} \right) \\
= \frac{3}{2} a .
\end{array}
$$

Therefore, $\frac{A D + B E + C F}{P D + P E + P F} = \sqrt{3}$.
Solution 2 Let point $P$ be the center of the equilateral $\triangle ABC$, and set $AB = 2 \sqrt{3}$. Then
$$
\begin{array}{l}
A D = B E = C F = \sqrt{3}, \\
P D = P E = P F = 1 .
\end{array}
$$

Hence, the answer is B.