Among the following, the pair of paramagnetic complexes is 

${\mathrm{K}}_3\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$

Central metal ion is ${\mathrm{Fe}}^{3+}$, which has $\left[\mathrm{Ar}\right]3{\mathrm{d}}^5$ configuration as free ion. In strong ligand field of ${\mathrm{CN}}^{-}$, its splits to become ${\mathrm{t}}_{2\mathrm{g}}{}^5\mathrm{e}_{\mathrm{g}}^0$.

One unpaired electron will be there, and 

$\therefore$ It is paramagnetic.

${\mathrm{K}}_3\left[{\mathrm{CoF}}_6\right]$ : Central metal ion is ${\mathrm{Co}}^{3+}$.

Free metal ion configuration $\left[\mathrm{Ar}\right]3{\mathrm{d}}^6$.

In weak ligand field of ${\mathrm{F}}^{-}$, its splits and becomes ${\mathrm{t}}_{2\mathrm{g}}^4{\mathrm{e}}_{\mathrm{g}}^2$.

Four unpaired electrons, therefore paramagnetic.

In ${\mathrm{K}}_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$, central ion is ${\mathrm{Fe}}^{2+}$ and in $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_3$, central ion is ${\mathrm{Co}}^{3+}$

In both of them, free metal ion is $\left(\mathrm{Ar}\right)3{\mathrm{d}}^6$.

Ligands ${\mathrm{NH}}_3$ and ${\mathrm{CN}}^{-}$ are of strong field, ${\mathrm{\Delta }}_0>\mathrm{P}$

So, ${\mathrm{d}}^6$ splits to ${\mathrm{t}}_{2\mathrm{g}}^6{\mathrm{e}}_{\mathrm{g}}^0$

No unpaired electrons, therefore diamagnetic.