Observe that $P$ is a truncated octahedron, formed by cutting off the corners from a regular octahedron with edge length 3. So, to compute the value of $P$, we can find the volume of the octahedron, and then subtract off the volume of truncated corners. Given a square pyramid where each triangular face an equilateral triangle, and whose side length is $s$, the height of the pyramid is $\frac{\sqrt{2}}{2}s$, and thus the volume is $\frac{1}{3} \cdot s^{2} \cdot \frac{\sqrt{2}}{2}s=\frac{\sqrt{2}}{6}s^{3}$. The side length of the octahedron is 3, and noting that the octahedron is made up of two square pyramids, its volume must be is $2 \cdot \frac{\sqrt{2}(3)^{3}}{6}=9\sqrt{2}$. The six "corners" that we remove are all square pyramids, each with volume $\frac{\sqrt{2}}{6}$, and so the resulting polyhedron $P$ has volume $9\sqrt{2}-6 \cdot \frac{\sqrt{2}}{6}=8\sqrt{2}$. Finally, to find the volume of all points closer to one particular vertex than any other vertex, note that due to symmetry, every point in $P$ (except for a set with zero volume), is closest to one of the 24 vertices. Due to symmetry, it doesn't matter which $V$ is picked, so we can just divide the volume of $P$ by 24 to obtain the answer $\frac{\sqrt{2}}{3}$.