Among the following complexes $\mathrm{K}_3[\mathrm{Fe}(\mathrm{CN})_6](\mathrm{K}),[\mathrm{Co}(\mathrm{NH}_3)_6]\mathrm{Cl}_3(\mathrm{L}),$ $\mathrm{Na}_3[\mathrm{Co}(\mathrm{oxalate})_3]$ (M), $[\mathrm{Ni}(\mathrm{H}_2\mathrm{O})_6]\mathrm{Cl}_2(\mathrm{N})$ $\mathrm{K}_2[\mathrm{Pt}(\mathrm{CN})_4]$ (O), and $\mathrm{Zn(H_2O)_6}_2$ (P) The diamagnetic complexes are

Check the metal oxidation state and $d$-electron count for each complex. For $\mathrm{K}_3[\mathrm{Fe}(\mathrm{CN})_6]$, the ion is $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$, so iron is $+3$ and hence $d^5$. Since $\mathrm{CN^-}$ is a strong-field ligand, it gives low-spin $t_{2g}^5$, which has one unpaired electron. So $\mathrm{K}$ is paramagnetic. For $[\mathrm{Co}(\mathrm{NH}_3)_6]\mathrm{Cl}_3$, the complex ion is $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$, so cobalt is $+3$ and hence $d^6$. With $\mathrm{Co^{3+}}$ and $\mathrm{NH_3}$, the octahedral complex is low spin $t_{2g}^6$, so all electrons are paired. Thus $\mathrm{L}$ is diamagnetic. For $\mathrm{Na}_3[\mathrm{Co}(\mathrm{oxalate})_3]$, the ion is $[\mathrm{Co}(\mathrm{oxalate})_3]^{3-}$. Since oxalate is $2-$, cobalt is $+3$, so it is $d^6$. This gives a paired configuration here, so $\mathrm{M}$ is diamagnetic. For $[\mathrm{Ni}(\mathrm{H}_2\mathrm{O})_6]\mathrm{Cl}_2$, the complex ion is $[\mathrm{Ni}(\mathrm{H}_2\mathrm{O})_6]^{2+}$, so nickel is $+2$ and hence $d^8$. Octahedral $\mathrm{Ni^{2+}}$ has two unpaired electrons, so $\mathrm{N}$ is paramagnetic. For $\mathrm{K}_2[\mathrm{Pt}(\mathrm{CN})_4]$, the ion is $[\mathrm{Pt}(\mathrm{CN})_4]^{2-}$, so platinum is $+2$ and hence $d^8$. Square-planar $d^8$ platinum complexes are diamagnetic. Thus $\mathrm{O}$ is diamagnetic. For $\mathrm{Zn(H_2O)_6}_2$, zinc is $+2$ with configuration $d^{10}$, so all electrons are paired. Thus $\mathrm{P}$ is diamagnetic. Therefore the diamagnetic complexes are $\mathrm{L, M, O, P}$. Comparing with the given options, this matches option $\mathrm{B}$.