Preparation of Amines

$CH_3CONH_2\xrightarrow[(ii)\ H_3O^{\oplus}]{(i)\ LiAlH_4}CH_3CH_2NH_2$ $CH_3CONH_2\xrightarrow{Br_2/KOH}CH_3NH_2$ $CH_3CN\xrightarrow[(ii)\ H_3O^{\oplus}]{(i)\ LiAlH_4}CH_3CH_2NH_2$ $CH_3NC\xrightarrow[(ii)\ H_3O^{\oplus}]{(i)\ LiAlH_4}CH_3NHCH_3$ Methyl isocyanide gives a secondary amine, $CH_3NHCH_3$, upon reduction.

In chlorobenzene, resonance gives the C–Cl bond partial double bond character, so aryl halides do not undergo nucleophilic substitution with phthalimide anion.

So Gabriel phthalimide synthesis cannot prepare aniline.

Work backwards from the product. $Br_2$/KOH converts an amide into a primary amine having one carbon less (Hoffmann bromamide degradation).
Ethylamine, $CH_3CH_2NH_2$, has two carbons, so the amide C has three carbons: C = $CH_3CH_2CONH_2$ (propanamide).
$CH_3CH_2CONH_2 + Br_2 + 4KOH \rightarrow CH_3CH_2NH_2 + K_2CO_3 + 2KBr + 2H_2O$
An amide is formed on heating the ammonium salt of the acid: B = $CH_3CH_2COO^-NH_4^+$ (ammonium propanoate), and $CH_3CH_2COONH_4 \xrightarrow{\Delta} CH_3CH_2CONH_2 + H_2O$
The ammonium salt is formed from the carboxylic acid and ammonia: $CH_3CH_2COOH + NH_3 \rightarrow CH_3CH_2COONH_4$
Hence A is propanoic acid, $CH_3CH_2COOH$.
Note: the question sentence is jumbled in the source PDF; it has been restored to its intended wording.

Step 1: thionyl chloride converts the carboxylic acid into the acid chloride: m-$BrC_6H_4COOH + SOCl_2 \rightarrow$ m-$BrC_6H_4COCl$ (B) $+ SO_2 + HCl$
Step 2: ammonia converts the acid chloride into the amide: m-$BrC_6H_4COCl + NH_3 \rightarrow$ m-$BrC_6H_4CONH_2$ (C) $+ HCl$
Step 3: with $Br_2$/NaOH the amide undergoes Hoffmann bromamide degradation, giving a primary amine with one carbon less: m-$BrC_6H_4CONH_2 + Br_2 + 4NaOH \rightarrow$ m-$BrC_6H_4NH_2$ (D) $+ Na_2CO_3 + 2NaBr + 2H_2O$
The ring bromine is not affected in any step.
Hence D is m-bromoaniline.

An amide treated with bromine and aqueous or ethanolic NaOH gives a primary amine with one carbon atom less (Hoffmann bromamide degradation).
$CH_3CONH_2 + Br_2 + 4NaOH \rightarrow CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
Acetamide has two carbons; the carbonyl carbon is lost as carbonate and the product is methyl amine.
$PCl_5$ dehydrates acetamide to acetonitrile ($CH_3CN$); soda lime and hot conc. $H_2SO_4$ do not give methyl amine (the latter hydrolyses it to acetic acid).
Hence the reagent is $NaOH-Br_2$.

Reduction of alkyl isocyanides in the presence of $LiAlH_4$ yields secondary amines containing methyl as one of the alkyl groups.
$R-N\equiv C + 4[H] \xrightarrow{LiAlH_4} R-NH-CH_3$ ($2^\circ$ amine)
For methyl isocyanide: $CH_3-N\equiv C + 4[H] \xrightarrow{LiAlH_4} CH_3-NH-CH_3$ (dimethylamine)
The carbon of the isocyanide group is attached to nitrogen, so after reduction it becomes a methyl group on nitrogen, giving a secondary amine.
Alkyl cyanides give $1^\circ$ amines on reduction: $CH_3-C\equiv N + 4[H] \rightarrow CH_3CH_2NH_2$
Acetamide also gives a primary amine: $CH_3CONH_2 + 4[H] \rightarrow CH_3CH_2NH_2 + H_2O$
Nitroethane gives the primary amine $CH_3CH_2NH_2$.
So the compound that yields a secondary amine is methyl isocyanide.