To determine which of the complexes are diamagnetic, we need to find the number of unpaired electrons in the d-orbitals of the central metal ion in each complex. A complex is diamagnetic if it has zero unpaired electrons. This involves determining the oxidation state of the metal, its d-electron configuration, the geometry of the complex, and the effect of the ligands (strong-field or weak-field) on the electron arrangement.

Let's analyze each complex:

1. K: $K_3[Fe(CN)_6]$

• Central metal ion: Iron (Fe).
• Oxidation state: The complex ion is $[Fe(CN)_6]^{3-}$. Let the oxidation state of Fe be $x$. Cyanide ($CN^-$) has a charge of -1. So, $x + 6(-1) = -3$, which gives $x = +3$. The ion is $Fe^{3+}$.
• Electron configuration: Fe (Z=26) is $[Ar] 3d^6 4s^2$. So, $Fe^{3+}$ is $[Ar] 3d^5$.
• Ligand and Geometry: $CN^-$ is a strong-field ligand. The geometry is octahedral. For a $d^5$ ion with a strong-field ligand, the complex is low-spin. The electrons will pair up in the lower energy $t_{2g}$ orbitals first.
• d-orbital filling: The configuration is $t_{2g}^5 e_g^0$. The five electrons in the three $t_{2g}$ orbitals will be arranged as $(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow)$.
• Conclusion: There is one unpaired electron. The complex is paramagnetic.

2. L: $[Co(NH_3)_6]Cl_3$

• Central metal ion: Cobalt (Co).
• Oxidation state: The complex ion is $[Co(NH_3)_6]^{3+}$. Ammonia ($NH_3$) is a neutral ligand. So, the oxidation state of Co is +3. The ion is $Co^{3+}$.
• Electron configuration: Co (Z=27) is $[Ar] 3d^7 4s^2$. So, $Co^{3+}$ is $[Ar] 3d^6$.
• Ligand and Geometry: $NH_3$ acts as a strong-field ligand with $Co^{3+}$. The geometry is octahedral. The complex is low-spin.
• d-orbital filling: The configuration is $t_{2g}^6 e_g^0$. The six electrons fill the three $t_{2g}$ orbitals, pairing up completely: $(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)$.
• Conclusion: There are zero unpaired electrons. The complex is diamagnetic.

3. M: $Na_3[Co(oxalate)_3]$

• Central metal ion: Cobalt (Co).
• Oxidation state: The complex ion is $[Co(C_2O_4)_3]^{3-}$. Oxalate ($C_2O_4^{2-}$) has a charge of -2. Let the oxidation state of Co be $x$. So, $x + 3(-2) = -3$, which gives $x = +3$. The ion is $Co^{3+}$.
• Electron configuration: $Co^{3+}$ is $[Ar] 3d^6$.
• Ligand and Geometry: Oxalate is a bidentate ligand, so the geometry is octahedral. For $Co^{3+}$, most ligands (including oxalate) cause electron pairing, leading to a low-spin complex.
• d-orbital filling: The configuration is $t_{2g}^6 e_g^0$. All six electrons are paired in the $t_{2g}$ orbitals.
• Conclusion: There are zero unpaired electrons. The complex is diamagnetic.

4. N: $[Ni(H_2O)_6]Cl_2$ (assuming the common hexaaqua complex, as $[Ni(H_2O)_3]Cl_2$ is an unusual representation for a monomeric species).

• Central metal ion: Nickel (Ni).
• Oxidation state: The complex ion is $[Ni(H_2O)_6]^{2+}$. Water ($H_2O$) is a neutral ligand. So, the oxidation state of Ni is +2. The ion is $Ni^{2+}$.
• Electron configuration: Ni (Z=28) is $[Ar] 3d^8 4s^2$. So, $Ni^{2+}$ is $[Ar] 3d^8$.
• Ligand and Geometry: $H_2O$ is a weak-field ligand. The geometry is octahedral. The complex is high-spin.
• d-orbital filling: The configuration is $t_{2g}^6 e_g^2$. The electrons fill the orbitals as $(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)$ in $t_{2g}$ and $(\uparrow)(\uparrow)$ in $e_g$.
• Conclusion: There are two unpaired electrons. The complex is paramagnetic.

5. O: $K_2[Pt(CN)_4]$

• Central metal ion: Platinum (Pt).
• Oxidation state: The complex ion is $[Pt(CN)_4]^{2-}$. So, $x + 4(-1) = -2$, which gives $x = +2$. The ion is $Pt^{2+}$.
• Electron configuration: Pt (Z=78) is a 5d element. Its configuration is $[Xe] 4f^{14} 5d^9 6s^1$. So, $Pt^{2+}$ is $[Xe] 4f^{14} 5d^8$.
• Ligand and Geometry: $CN^-$ is a strong-field ligand. For $d^8$ ions of 4d and 5d series, square planar geometry is highly favored. This leads to a low-spin configuration.
• d-orbital filling: In a square planar field, the 8 electrons fill the lower four d-orbitals ($d_{xz}, d_{yz}, d_{z^2}, d_{xy}$), leaving the high-energy $d_{x^2-y^2}$ orbital empty. All electrons are paired.
• Conclusion: There are zero unpaired electrons. The complex is diamagnetic.

6. P: $[Zn(H_2O)_6](NO_3)_2$ (assuming standard notation where nitrate is the counter-ion).

• Central metal ion: Zinc (Zn).
• Oxidation state: The complex ion is $[Zn(H_2O)_6]^{2+}$. So, the oxidation state of Zn is +2. The ion is $Zn^{2+}$.
• Electron configuration: Zn (Z=30) is $[Ar] 3d^{10} 4s^2$. So, $Zn^{2+}$ is $[Ar] 3d^{10}$.
• Ligand and Geometry: The d-subshell is completely filled.
• d-orbital filling: A $d^{10}$ configuration means all d-orbitals are filled with paired electrons, regardless of the ligand field strength or geometry.
• Conclusion: There are zero unpaired electrons. The complex is diamagnetic.

Summary: The diamagnetic complexes are L, M, O, and P. This corresponds to option C.