The hybridization and magnetic nature of [Mn(CN)₆]^4– and [Fe(CN)₆]^3–respectively are:

CN is strong field ligand it involves inner d- orbital ${\mathrm{Mn}}^{+2}$${\mathrm{t}}_{2\mathrm{g}}^4{\mathrm{e}}_{\mathrm{g}}^0$   ;  ${\mathrm{Fe}}^{+3}$  ${\mathrm{t}}_{2\mathrm{g}}^4{\mathrm{e}}_{\mathrm{g}}^0$ hence 2 unpaired electron in both.

Hence, Option B is correct.