 Solution

Without a blocking diphosphine, square-planar Pd(II) can bind four terminal pyridines. If there are $N$ Fe units and $M$ Pd centres, saturation requires $3N=4M$. The given ratio $M/N=0.75$ and charge $2(N+M)=28$ yield $N=8$, $M=6$. Thus $[{\mathrm{Fe}}_8{\mathrm{Pd}}_6{L}_8{]}^{28+}$ has Fe at the eight cube corners and Pd at the six face centres. Each corner Fe connects through its three arms to the three neighbouring face-centre Pd atoms; each Pd accepts four pyridines around its square face.

There are 24 Fe-unit-to-Pd connections, checked independently as $8\times 3=6\times 4$. It is useful to distinguish the cube of eight Fe nodes from its dual octahedron of six Pd nodes: both describe the same assembly, with different node sets.