1. Identify compound A from molecular formula and test

   The compound has formula $C_6H_6O$ and gives a dark green colour with ferric chloride.

   • Ferric chloride test indicates the presence of a phenolic $-OH$ group.
   • The compound with formula $C_6H_6O$ containing a phenolic group is phenol: $C_6H_5OH$

   So, A = phenol.

2. Reaction of phenol with $CHCl_3/KOH$ followed by acidification

   This is the Reimer–Tiemann reaction.

   Phenol on treatment with $CHCl_3$ and $KOH$, followed by acidification, gives mainly o-hydroxybenzaldehyde (salicylaldehyde).

   $C_6H_5OH \xrightarrow[\text{acidification}]{CHCl_3/KOH} o\text{-}HOC_6H_4CHO$

   So, B = salicylaldehyde (2-hydroxybenzaldehyde).

3. Identify compound C such that PCC gives B

   PCC oxidizes a primary alcohol to an aldehyde without further oxidation to acid.

   Since B is $o$-hydroxybenzaldehyde, the corresponding alcohol C must be:

   $o\text{-}HOC_6H_4CH_2OH$

   which is 2-hydroxybenzyl alcohol (salicyl alcohol).

   $o\text{-}HOC_6H_4CH_2OH \xrightarrow{PCC} o\text{-}HOC_6H_4CHO$

   So, C = 2-hydroxybenzyl alcohol.

4. Final identification

   • $A =$ Phenol
   • $B =$ o-Hydroxybenzaldehyde (salicylaldehyde)
   • $C =$ o-Hydroxybenzyl alcohol

5. Match with options

   This set corresponds to Option A.

Therefore, the correct answer is A.