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Best for you to try and draw a picture of this:

$P$ is a point on the line segment $AB$. Meaning that the "triangle" $ABP$, having perimeter line segments $AB$, $BP$ and $PA$, is just the line segment $AB$ itself (since all 3 of the perimeter line segments are subsets of $AB$).

Now we can say that the area of $ABP$ equals the area of $AB$, which is $0$ because the area of a line segment is $0$ (line segments have effectively $0$ width, so $area=length\times 0=0$)