Let $\triangle ABC$ be a given triangle. For any point $P$ inside this triangle, define the intersections $A(P), B(P), C(P)$ as follows:

- $A(P)$ is the intersection of line $AP$ with side $BC$.
- $B(P)$ is the intersection of line $BP$ with side $CA$.
- $C(P)$ is the intersection of line $CP$ with side $AB$.

We aim to determine the position of $P$ such that the area of $\triangle A(P)B(P)C(P)$ is maximized.

### Analyzing the Geometry

The area of $\triangle A(P)B(P)C(P)$ is closely tied to the location of $P$. Specifically, this area is maximized when $P$ is the centroid of $\triangle ABC$. This conclusion can be drawn by considering the specific properties of the centroid:

- The centroid divides each median in a 2:1 ratio.
- It is the point within $\triangle ABC$ where the triangle is divided into smaller triangles of equal area.

### Area Calculation

For the maximal area condition, consider $P$ to be the centroid $G$ of $\triangle ABC$. The area of the triangle $\triangle A(P)B(P)C(P)$ formed by the cevians (medians) is known from the properties of centroids:

$$\text{Area of } \triangle A(P)B(P)C(P) = \frac{1}{4} \times \text{Area of } \triangle ABC$$

This formula arises from the fact that the centroid divides the triangle into smaller triangles each having equal area, resulting in four smaller triangles each having one-fourth the area of $\triangle ABC$.

### Conclusion

Thus, when $P$ is placed at the centroid of the triangle $\triangle ABC$, the area of triangle $\triangle A(P)B(P)C(P)$ becomes:

\[
\boxed{\frac{S_{\triangle ABC}}{4}}
]

This completes the solution for determining $P$ such that the area of $\triangle A(P)B(P)C(P)$ is maximized.