10. Given that the four vertices of the tetrahedron $P-ABC$ lie on the surface of sphere $O$, $PA=PB=PC$. $\triangle ABC$ is an equilateral triangle with side length 2, $E$ and $F$ are the midpoints of $AC$ and $BC$ respectively, $\angle EPF=60^{\circ}$. Then the surface area of sphere $O$ is
10. $6 \pi$.

Given that $P-ABC$ is a regular tetrahedron, we know $EP=FP$. Therefore, $\triangle EPF$ is an equilateral triangle with a side length of 1.
From this, we know that the height of the lateral face $PAC$ is $PE=1$.
Thus, the edge length $PA=\sqrt{2}$.
By restoring it to a cube with an edge length of $\sqrt{2}$, we find that the diameter of the circumscribed sphere of the regular tetrahedron $P-ABC$ is exactly the body diagonal of the cube, which is $\sqrt{6}$. Therefore, the surface area is $6 \pi$.