1. Identify the starting compound and target product

Cyanobenzene is benzonitrile, with structure: $C_6H_5-C\equiv N$

The target product is benzylamine: $C_6H_5-CH_2NH_2$

So we need to convert a nitrile into a primary amine.

A nitrile on reduction gives: $R-C\equiv N \xrightarrow[\text{reduction}]{} R-CH_2NH_2$

2. Check each option

Option B: $H_2/Ni$

Catalytic hydrogenation of nitriles reduces them to primary amines: $C_6H_5-C\equiv N + 2H_2 \xrightarrow{Ni} C_6H_5-CH_2NH_2$ So B is a correct method.

Option C: (i) $LiAlH_4$ (ii) $H_3O^+$

$LiAlH_4$ is a standard reagent for reducing nitriles to primary amines: $C_6H_5-C\equiv N \xrightarrow[2.\ H_3O^+]{1.\ LiAlH_4} C_6H_5-CH_2NH_2$ So C is a correct method.

Option A: (i) $SnCl_2 + HCl$(gas) (ii) $NaBH_4$

This is Stephen reduction. A nitrile is first reduced to an iminium salt, which on hydrolysis gives an aldehyde: $C_6H_5-C\equiv N \xrightarrow{SnCl_2/HCl} \text{iminium salt} \xrightarrow{H_2O} C_6H_5-CHO$ Then $NaBH_4$ reduces benzaldehyde to benzyl alcohol, not benzylamine: $C_6H_5-CHO \xrightarrow{NaBH_4} C_6H_5-CH_2OH$ Therefore this sequence does not give benzylamine.

So A is not a correct method.

Option D: (i) $HCl/H_2O$ (ii) $NaBH_4$

Acidic hydrolysis of benzonitrile gives benzoic acid: $C_6H_5-C\equiv N \xrightarrow{HCl/H_2O} C_6H_5-COOH$ Now $NaBH_4$ is a mild reducing agent. It reduces aldehydes and ketones, but does not reduce carboxylic acids under ordinary conditions. So benzoic acid will not be converted to benzylamine.

Hence D is also not a correct method.

3. Conclusion

Among the given options, A and D both are not correct methods for preparing benzylamine from cyanobenzene.

Since the question is marked as MCQ (single correct) but two options are incorrect, the stored answer appears incomplete/incorrect.

Therefore, the chemically correct conclusion is: $\boxed{A \text{ and } D}$