The following structures are

First write down the four groups that are bonded to the tetrahedral carbon in each drawing: $$Br$$, $$Cl$$, $$Me$$ and $$CH_3$$.

The symbols $$Me$$ and $$CH_3$$ mean exactly the same thing - a methyl group. Hence the carbon is attached to only three different groups:

$$Br,\;Cl,\;\text{and a methyl group (present twice)}$$

A carbon atom becomes a stereogenic (chiral) centre only when all four substituents are different. Because two of the substituents here are identical, the carbon is achiral; it cannot possess the $$R/S$$ configuration and cannot have any non-superimposable mirror image.

Therefore any two drawings that place these same groups around that carbon—no matter how the wedges and dashes are arranged—represent the same achiral molecule. They are not enantiomers (mirror images that are non-superimposable), not diastereomers (stereoisomers that are not mirror images) and not meso forms (which require at least two chiral centres with internal compensation).

Hence the two structures are identical molecules.

Option B which is: Identical