【Analysis】
Draw lines parallel to the three sides through point $P$, thus dividing the figure into three parallelograms, $\square A G P H, \square B K P J$, $\square C M P L$, the shaded area of each occupies $\frac{1}{2}$ of its own; similarly, the three triangles $\triangle H J P, \triangle L P K, \triangle M P G$, the shaded area of each also occupies $\frac{1}{2}$ of its own; therefore, $S_{\triangle P A D}+S_{\triangle P B E}+S_{\triangle P C F}=\frac{1}{2} S_{\triangle A B C}$, solving based on this.

【Solution】Solution: According to the analysis,
$$
S_{\triangle P A D}+S_{\triangle P B E}+S_{\triangle P C F}=\frac{1}{2} S_{\triangle A B C},
$$

Given that the area of the equilateral triangle $A B C$ is 2028 square centimeters, and the areas of triangles $P A D$ and $P B E$ are both 192 square centimeters,

Therefore, $S_{\triangle P C F}=2028 \times \frac{1}{2}-192 \times 2$
$$
=1014-384
$$
$=630$ (square centimeters)
Answer: The area of triangle $P C F$ is 630 square centimeters.
The answer is: 630.
8. (10 points) Through a point $P$ inside an equilateral triangle $A B C$, perpendiculars are drawn to the three sides, with the feet of the perpendiculars being $D, E, F$ respectively. Connect $A P$, $B P$, $C P$. If the area of the equilateral triangle $A B C$ is 2028 square centimeters, and the areas of triangles $P A D$ and $P B E$ are both 192 square centimeters, then the area of triangle $P C F$ is $\qquad$ square centimeters.