Solution:
Let the line divide $P$ into two polygons $P_1$ and $P_2$ with equal areas and equal perimeters. Suppose, for contradiction, that the dividing line $\ell$ does not pass through $O$.

Let $r$ be the radius of the inscribed circle. The inscribed circle is tangent to every side of $P$.

Let $d$ be the (signed) distance from $O$ to $\ell$. Consider the two regions $P_1$ and $P_2$.

The perimeter of $P$ is equal to the sum of the lengths of the tangents from $O$ to the sides, which is $n r$ for an $n$-gon, but more generally, the sum of the distances from $O$ to the sides times the number of sides.

However, more simply, note that the inscribed circle is tangent to every side, so the sum of the distances from $O$ to the sides (counted with multiplicity) is constant.

Now, if the line $\ell$ does not pass through $O$, then the two regions $P_1$ and $P_2$ have boundaries consisting of parts of the boundary of $P$ and the segment of $\ell$ inside $P$.

The perimeters of $P_1$ and $P_2$ are each equal to the sum of the lengths of their respective portions of the boundary of $P$ plus the length of the segment of $\ell$ inside $P$.

But the sum of the perimeters of $P_1$ and $P_2$ is equal to the perimeter of $P$ plus twice the length of the segment of $\ell$ inside $P$ (since this segment is counted in both perimeters).

If the perimeters of $P_1$ and $P_2$ are equal, then each is equal to half the sum, so each is greater than half the perimeter of $P$ unless the segment of $\ell$ inside $P$ has zero length, i.e., unless $\ell$ passes through $O$ (the center of the inscribed circle), in which case the two regions are congruent in terms of their tangency to the circle.

Alternatively, consider the following argument:

Suppose the line $\ell$ does not pass through $O$. Then the two regions $P_1$ and $P_2$ are not symmetric with respect to $O$. The inscribed circle touches the boundary of $P$ at points equidistant from $O$. The only way for the two regions to have equal areas and equal perimeters is for the line to pass through $O$, so that the two regions are mirror images with respect to $O$.

Therefore, the dividing line must pass through $O$.