$$[Pd(F)(Cl)(Br)(I)]^{2-}$$ 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. 

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.