The tetrahedra ${\left[{\mathrm{CoI}}_4\right]}^{2-}$ and square planar ${\left[{\mathrm{PdBr}}_4\right]}^2$  complex ions are respectively :

Since ${\left[{\mathrm{CoI}}_4\right]}^{2-}$ has a greater number of unpaired electrons it has high spin while ${\left[{\mathrm{PdBr}}_4\right]}^{2-}$ has low spin.

Hybridization of the tetrahedral ${\left[{\mathrm{CoI}}_4\right]}^{2-},{\mathrm{Co}}^{+2},{\mathrm{d}}^7,{\mathrm{sp}}^3$ atom is highly difficult.

Low spin complex square planar ${\left[{\mathrm{PdBr}}_4\right]}^{2-},{\mathrm{Pd}}^{+2},{\mathrm{d}}^8,{\mathrm{dsp}}^2$ hybridization.

Hence, option B is correct.