By symmetry each octahedron vertex is the same distance $t$ from $P_1$ (or $P_1'$) along its edge

Set the squared distance between two vertices sharing $P_1$ equal to the squared distance across, and solve for $t$

Place $P_1$ at the origin with edges along the axes, and let each of the three octahedron vertices near $P_1$ be a distance $t$ from $P_1;$ by symmetry the three near $P_1'$ are also a distance $t$ from $P_1'.$

Two vertices sharing $P_1,$ such as $(t,0,0)$ and $(0,t,0),$ are a distance $t\sqrt2$ apart. A vertex near $P_1,$ say $(t,0,0),$ and the appropriate vertex near $P_1',$ say $(1,1-t,1),$ must be the same distance apart.

Setting the two squared side lengths equal and using the cube’s unit edges yields $t=\tfrac34,$ so the side length is $t\sqrt2=\dfrac{3\sqrt2}{4}.$

Thus, the correct answer is A.