Reaction of Grignard reagent, C$$_2$$H$$_5$$MgBr with C$$_8$$H$$_8$$O followed by hydrolysis gives compound A, which reacts instantly with Lucas reagent to give compound B(C$$_{10}$$H$$_{13}$$Cl). The Compound B is:

The Grignard reagent supplied is $$C_2H_5MgBr$$ (ethyl-magnesium bromide). Grignard reagents add to carbonyl compounds to give alcohols after hydrolysis.

The unknown carbonyl compound has molecular formula $$C_8H_8O$$. A well-known aromatic ketone with this formula is acetophenone, $$C_6H_5COCH_3$$ (phenyl methyl ketone).

Step-1 (Grignard addition):
$$C_6H_5COCH_3 + C_2H_5MgBr \;\xrightarrow[]{\phantom{\,}}\; C_6H_5C\big(O^{-}MgBr\big)(CH_3)(C_2H_5)$$

Step-2 (acidic hydrolysis):
$$C_6H_5C\big(O^{-}MgBr\big)(CH_3)(C_2H_5) \;+\; H_3O^{+}\;\rightarrow\;C_6H_5C(OH)(CH_3)(C_2H_5)\;+\;Mg(OH)Br$$

The product $$A$$ is therefore the tertiary alcohol
$$\text{Ph-C(OH)(CH}_3)(C_2H_5)$$
Its molecular formula is $$C_{10}H_{14}O$$ (10 C from Ph (6) + quaternary C(1) + CH_3(1) + C_2H_5(2); 14 H from Ph (5) + CH_3(3) + C_2H_5(5) + OH (1)).

Lucas test: An alcohol that gives an immediate cloudy layer with Lucas reagent (conc. $$HCl/ZnCl_2$$) must be tertiary, proceeding by the $$S_N1$$ route.

Substitution of the $$OH$$ group in $$A$$ by $$Cl^{-}$$ gives compound $$B$$:
$$\text{Ph-C(OH)(CH}_3)(C_2H_5)\;+\;HCl \;\xrightarrow[ZnCl_2]{}\; \text{Ph-C(Cl)(CH}_3)(C_2H_5)\;+\;H_2O$$

Structure of $$B$$: $$\text{Ph-C(Cl)(CH}_3)(C_2H_5)$$

Counting atoms: Carbons 6 (phenyl) +1 (quaternary C)+1 (CH_3)+2 (C_2H_5)=10; Hydrogens 5 (phenyl)+3+5=13; one chlorine atom. Thus molecular formula $$C_{10}H_{13}Cl$$ matches the data.

Systematic name: $$2$$-chloro-$$2$$-phenylbutane (equivalently $$1$$-chloro-$$1$$-phenylbutane in common nomenclature).

Option 3 represents the structure $$\text{Ph-C(Cl)(CH}_3)(C_2H_5)$$, so Option 3 is the correct choice for compound $$B$$.