4. As shown in Figure 3, in $\triangle A B C$, let $E$, $F$, and $P$ be on sides $B C$, $C A$, and $A B$ respectively. Given that $A E$, $B F$, and $C P$ intersect at a point $D$, and $\frac{A D}{D E}+\frac{B D}{D F}+\frac{C D}{D P}=n$. Then $\frac{A D}{D E} \cdot \frac{B D}{D F} \cdot \frac{C D}{D P}=$
\begin{array}{l}\text { 4. } n+2 \text {. } \\ \text { Let } S_{\triangle B D C}=x, S_{\triangle C D A}=y, S_{\triangle A D B}=z \text {. Then } \\ \frac{A D}{D E}=\frac{z+y}{x}, \frac{B D}{D F}=\frac{x+z}{y}, \frac{C D}{D P}=\frac{x+y}{z} . \\ \text { Therefore, } \frac{A D}{D E} \cdot \frac{B D}{D F} \cdot \frac{C D}{D P} \\ =\frac{(z+y)(x+z)(x+y)}{x y z} \\ =\frac{z+y}{x}+\frac{x+z}{y}+\frac{x+y}{z}+2 \\ =\frac{A D}{D E}+\frac{B D}{D F}+\frac{C D}{D P}+2 \\ =n+2 .\end{array}