$Al{F}_3$ is soluble in $HF$ only in presence of $KF$. It is due to the formation of

The correct option is B $K_3\left[Al{F}_6\right]$
$Al{F}_3$ is insoluble in anhydrous $HF$ because the $F^{-}$ ions are not available inhydrogen bonded $HF$ molecules but, it becomes soluble in presence of little amount of $KF$ due to formation of complex $K_3\left[Al{F}_6\right]$
$Al{F}_3+3KF\to K_3\left[Al{F}_6\right]$