The hybridization of Fe in $\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6]$ complex is

In $\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6]$, the complex ion is $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$. Since each $\mathrm{CN}^-$ ligand has charge $-1$, the oxidation state of iron is $+2$. For $\mathrm{Fe}^{2+}$, the electronic configuration is $3d^6$. $\mathrm{CN}^-$ is a strong field ligand, so it causes pairing of electrons in the $3d$ orbitals, giving a low-spin octahedral complex. Thus two $3d$ orbitals, one $4s$, and three $4p$ orbitals hybridize. So the hybridization is $\mathrm{d}^2\mathrm{sp}^3$. Comparing with the given options, this matches option $\mathrm{d}^2\mathrm{sp}^3$, so the correct choice is $C$.