The molar ionic conductance of octahedral complexes is given below : 

(P) ${\mathrm{PtCl}}_4\cdot 5{\mathrm{NH}}_3$      (Q) ${\mathrm{PtCl}}_4\cdot 4{\mathrm{NH}}_3$

(R) ${\mathrm{PtCl}}_4\cdot 3{\mathrm{NH}}_3$       (S) ${\mathrm{PtCl}}_4\cdot 2{\mathrm{NH}}_3$

Correct order is :

 

Since the number free ions increase the molar ionic conductance the correct order of molar ionic  conductance is $\mathrm{P}>\mathrm{Q}>\mathrm{R}>\mathrm{S}$

The number of free ions for the following compound is,

$$\begin{gathered} \left(\mathrm{P}\right) \left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_5\mathrm{Cl}\right]{\mathrm{Cl}}_3\Rightarrow 4\text{ ions } \\ \left(\mathrm{Q}\right) \left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2\right]{\mathrm{Cl}}_2\Rightarrow 3\text{ ions } \\ \left(\mathrm{R}\right) \left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_3{\mathrm{Cl}}_3\right]\mathrm{Cl}\Rightarrow 2\text{ ions } \\ \left(\mathrm{S}\right) \left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_2{\mathrm{Cl}}_4\right]\Rightarrow 0\text{ ions } \end{gathered}$$

Hence correct option is B.