$$[Pd(F)(Cl)(Br)(I)]^{2-}$$ has $$n$$ number of geometrical isomers. Then, the spin-only magnetic moment and crystal field stabilization energy [CFSE] of $$[Fe(CN)_6]^{n-6}$$, respectively, are:
[Note: Ignore the pairing energy]

We begin with the square-planar complex $$[Pd(F)(Cl)(Br)(I)]^{2-}$$. In a square plane the four positions are equivalent by rotation, so one ligand may be fixed at the “north” position without loss of generality. After fixing one ligand, only the relative arrangement of the remaining three ligands decides whether two drawings are identical or different. Exhaustive placement (or the standard group-theoretical count) shows that a square-planar complex that contains four different ligands (type $$MABCD$$) possesses exactly three distinct geometrical isomers.

Hence the number of geometrical isomers is

$$n = 3.$$

The same symbol $$n$$ now appears in the charge of the next complex:

$$[Fe(CN)_6]^{\,n-6}.$$

Substituting $$n = 3$$ obtained above, we have

$$[Fe(CN)_6]^{\,3-},$$

so the oxidation state of iron is

$$x + 6(-1) = -3 \;\;\Longrightarrow\;\; x = +3.$$

Thus we are dealing with $$Fe^{3+}$$, which has the electronic configuration $$3d^5$$. Because the ligand $$CN^-$$ is a strong-field (spectrochemical series) ligand, the octahedral complex will be low spin, giving the distribution

$$t_{2g}^{\,5}\;e_g^{\,0}.$$

Number of unpaired electrons:

The five electrons fill the three $$t_{2g}$$ orbitals pairwise as far as possible, leaving only one orbital with a single electron. Therefore

$$n_{\text{unpaired}} = 1.$$

We now calculate the spin-only magnetic moment.
Formula stated first:

$$\mu_{\text{spin only}} = \sqrt{n(n+2)}\;\text{BM},$$

where $$n$$ is the number of unpaired electrons. Substituting $$n = 1$$ gives

$$\mu = \sqrt{1(1+2)} = \sqrt{3} \;\text{BM} \approx 1.73 \;\text{BM}.$$

Next, we evaluate the crystal-field stabilisation energy (CFSE) in an octahedral field, ignoring pairing energy as instructed.
General CFSE expression:

$$\text{CFSE} = (-0.4\Delta_0)\times N_{t_{2g}} + (+0.6\Delta_0)\times N_{e_g}.$$

For the configuration $$t_{2g}^{\,5}e_g^{\,0}$$,

$$\text{CFSE} = (-0.4\Delta_0)\times 5 + (+0.6\Delta_0)\times 0 = -2.0\Delta_0.$$

Thus, the spin-only magnetic moment is $$\approx 1.73\;\text{BM}$$ and the CFSE is $$-2.0\Delta_0$$.

Hence, the correct answer is Option C.