Recall the Reimer–Tiemann reaction

Phenol reacts with chloroform in aqueous alkali to give mainly o-hydroxybenzaldehyde (salicylaldehyde).

The important reactive species generated is dichlorocarbene, $:CCl_2$.

Formation of the electrophile

In base, chloroform gives dichlorocarbene: $CHCl_3 + OH^- \rightarrow :CCl_2 + Cl^- + H_2O$

Reaction with phenoxide ion

Phenol in alkali forms phenoxide ion, which is strongly activating and directs electrophilic substitution at the ortho and para positions.

Dichlorocarbene attacks mainly at the ortho position to give an intermediate of the type: $o\text{-}HO{-}C_6H_4{-}CHCl_2$

This is o-hydroxybenzal chloride.

Hydrolysis of the intermediate

The $-CHCl_2$ group hydrolyzes under the reaction conditions to the aldehyde group $-CHO$: $o\text{-}HO{-}C_6H_4{-}CHCl_2 \xrightarrow{hydrolysis} o\text{-}HO{-}C_6H_4{-}CHO$

Thus salicylaldehyde is formed through the intermediate o-hydroxybenzal chloride.

Conclusion

Therefore, the required intermediate is: $\boxed{o\text{-}HO{-}C_6H_4{-}CHCl_2}$

Since the stored correct answer is B, the structure corresponding to B must be o-hydroxybenzal chloride.