The species formed on fluorination of phosphorus pentachloride in a polar organic solvent are

Phosphorus pentachloride is a trigonal-bipyramidal molecule. In a highly polar organic solvent it undergoes auto-ionisation (self-dissociation) according to

$$PCl_5 \;\longrightarrow\; [PCl_4]^+ + [PCl_6]^- \quad -(1)$$

When the solution is treated with a fluoride ion source (for example $$HF$$, $$SbF_5$$, or any fluorinating agent), the incoming fluoride ions attack the halophosphate anion first because the anion already carries an excess negative charge and can accept further halide substitution more easily than the cation.

Step 1 - Replacement of two equatorial chlorides
Fluoride ions substitute two of the equatorial chlorides of $$[PCl_6]^-$$ to give the mixed-halide anion $$[PCl_4F_2]^-$$:

$$[PCl_6]^- + 2\,F^- \;\longrightarrow\; [PCl_4F_2]^- + 2\,Cl^- \quad -(2)$$

The cation produced in reaction $$(1)$$ remains unchanged, so the ionic species present after this first fluorination are

$$[PCl_4]^+[PCl_4F_2]^-$$

Step 2 - Complete fluorination
If further fluoride ions are supplied, all six positions around phosphorus in the anion can be occupied by fluorides, converting $$[PCl_4F_2]^-$$ into the hexafluorophosphate anion $$[PF_6]^-$$:

$$[PCl_4F_2]^- + 4\,F^- \;\longrightarrow\; [PF_6]^- + 4\,Cl^- \quad -(3)$$

The cation is still $$[PCl_4]^+$$, so a second ionic pair appears:

$$[PCl_4]^+[PF_6]^-$$

Thus, on fluorination of $$PCl_5$$ in a polar organic solvent, two distinct ion pairs can be isolated depending on how many fluorides replace chlorides in the anion:

$$[PCl_4]^+[PCl_4F_2]^- \quad\text{and}\quad [PCl_4]^+[PF_6]^-$$

None of the other suggested combinations can be obtained under these conditions because the cationic part remains $$[PCl_4]^+$$ and the neutral covalent molecules $$PF_5$$, $$PF_3$$ or $$PCl_3$$ are not formed in appreciable amounts in a polar medium.

Therefore, the correct choice is:
Option B which is: $$[PCl_4]^+[PCl_4F_2]^-$$ and $$[PCl_4]^+[PF_6]^-$$