5. (7 points) On side $AB$ of triangle $ABC$, a point $K$ is marked, and on side $AC$ - a point $M$. Segments $BM$ and $CK$ intersect at point $P$. It turns out that angles $APB$, $BPC$, and $CPA$ are each $120^{\circ}$, and the area of quadrilateral $AKPM$ is equal to the area of triangle $BPC$. Find the angle $BAC$.

#