Because $\triangle PAB$ has area $1/3$ of the area of $\triangle ABC$, the height of $\triangle PAB$, with respect to the base $AB$, is $1/3$ of the height of $\triangle ABC$ with respect to $AB$. Thus $P$ lies on the line $l$ parallel to $AB$, such that the distance between $l$ and $AB$ is $1/3$ of the distance from $C$ to $AB$.
For the same reason, $P$ lies on the line $m$ parallel to $BC$, such that the distance between $m$ and $BC$ is $1/3$ of the distance from $A$ to $BC$.
But because the centroid of $\triangle ABC$ divides each of the medians in the ratio $2:1$, it follows that the centroid also lies on both $l$ and $m$. Since there is only one point which is on both $l$ and $m$, the centroid must be $P$.