In the spectrochemical series, ligand field strength decreases in the order:

$$\text{CO} \gt \text{CN}^- \gt \text{en} \gt \text{NH}_3 \gt \text{NCS}^- \gt \text{H}_2\text{O} \gt \text{F}^- \gt \text{Cl}^- \gt \text{S}^{2-} \gt \text{Br}^- \gt \text{I}^-$$

Comparing the given four ligands:

$$\text{CN}^-\text{ (D)} \gt \text{en (B)} \gt \text{NCS}^-\text{ (C)} \gt \text{S}^{2-}\text{ (A)}$$

Decreasing order: (D) > (B) > (C) > (A).

In $[\text{Fe}(\text{CN})_6]^{3-}$:

• Iron is in $+3$ oxidation state: $\text{Fe}^{3+} = [\text{Ar}]3d^5$.
• Cyanide ($\text{CN}^-$) is a strong field ligand, causing pairing of electrons in $3d$ orbitals: configuration becomes $t_{2g}^5 e_g^0$ with $1$ unpaired electron.
• Because $n = 1$, the complex is Paramagnetic (Statement A is correct) and $\mu = \sqrt{1(3)} \approx 1.73\text{ BM}$ (not $5.92\text{ BM}$, so C is false).
• Two vacant inner $3d$ orbitals, along with one $4s$ and three $4p$ orbitals, undergo $d^2sp^3$ hybridization (Statement D is correct; B is false).

When $[\text{Cr}(\text{NH}_3)_6]\text{Cl}_3$ dissolves in water, the three chloride ions present outside the coordination sphere ionise completely:

$$[\text{Cr}(\text{NH}_3)_6]\text{Cl}_3 (aq) \rightarrow [\text{Cr}(\text{NH}_3)_6]^{3+} (aq) + 3\text{Cl}^- (aq)$$

This yields $1$ complex cation $[\text{Cr}(\text{NH}_3)_6]^{3+}$ and $3$ chloride anions $\text{Cl}^-$, giving a total of $1 + 3 = 4$ ions.