jmc

Since the three triangles $ABP$, $ACP$, and $BCP$ have equal bases, their areas are proportional to the lengths of their altitudes.

Let $O$ be the centroid of $△ABC$, and draw medians $\overline{AOE}$ and $\overline{BOD}$. Any point above $\overline{BOD}$ will be farther from $\overline{AB}$ than from $\overline{BC},$ and any point above $\overline{AOE}$ will be farther from $\overline{AB}$ than from $\overline{AC}.$ Therefore the condition of the problem is met if and only if $P$ is inside quadrilateral $CDOE.$



If $\overline{CO}$ is extended to $F$ on $\overline{AB}$, then $△ABC$ is divided into six congruent triangles, of which two comprise quadrilateral $CDOE$. Thus $CDOE$ has one-third the area of $△ABC,$ so the required probability is $\boxed{\frac{1}{3}}$.