1. Find oxidation state and electronic configuration of the metal in each complex

We check whether the complex has unpaired electrons.

2. Option A: $\mathrm{K_3[Cu(CN)_4]}$

Let oxidation state of Cu be $x$.

$x+4(-1)=-3$ $x-4=-3$ $x=+1$

So, Cu is $\mathrm{Cu^+}$.

Atomic number of Cu = 29

Neutral Cu: $[Ar]3d^{10}4s^1$

For $\mathrm{Cu^+}$: $[Ar]3d^{10}$

This is a completely filled $d$-subshell, so all electrons are paired. Hence, this complex is diamagnetic.

3. Option B: $\mathrm{K_2[Cu(CN)_4]}$

Let oxidation state of Cu be $x$.

$x+4(-1)=-2$ $x-4=-2$ $x=+2$

So, Cu is $\mathrm{Cu^{2+}}$.

For $\mathrm{Cu^{2+}}$: $[Ar]3d^9$

A $d^9$ configuration has one unpaired electron. So this complex is paramagnetic.

4. Option C: $\mathrm{K_3[Fe(CN)_4]}$

Let oxidation state of Fe be $x$.

$x+4(-1)=-3$ $x-4=-3$ $x=+1$

So, Fe is $\mathrm{Fe^+}$.

Atomic number of Fe = 26

Neutral Fe: $[Ar]3d^6 4s^2$

For $\mathrm{Fe^+}$: $[Ar]3d^7 4s^0 \text{ (effectively }3d^7\text{)}$

A $d^7$ configuration generally has unpaired electrons, so it is paramagnetic.

5. Option D: $\mathrm{K_4[FeCl_6]}$

Let oxidation state of Fe be $x$.

$x+6(-1)=-4$ $x-6=-4$ $x=+2$

So, Fe is $\mathrm{Fe^{2+}}$.

For $\mathrm{Fe^{2+}}$: $[Ar]3d^6$

$\mathrm{Cl^-}$ is a weak field ligand, so the complex is high spin. For octahedral high-spin $d^6$, there are 4 unpaired electrons. Hence it is paramagnetic.

6. Conclusion

Only Option A is diamagnetic.

$\boxed{\mathrm{K_3[Cu(CN)_4]}}$