1. The reagent $Br_2/NaOH$ indicates the Hofmann bromamide reaction.

   In this reaction, an amide of type $RCONH_2 \xrightarrow{Br_2/NaOH} RNH_2$ gives a primary amine with one carbon less.

2. The product formed is given as having molecular formula $C_3H_9N$ and it gives a positive carbylamine test.

   • Carbylamine test is given only by primary amines.
   • A primary amine with formula $C_3H_9N$ is propylamine $(C_3H_7NH_2)$.

   Therefore, the Hofmann reaction product must be a 3-carbon primary amine.

3. Hence the starting amide must have 4 carbons and be of the form: $C_3H_7CONH_2$

4. Now check the options:

   A. $CH_3CH_2CH_2CONH_2$

   • This is butanamide.
   • On Hofmann bromamide degradation: $CH_3CH_2CH_2CONH_2 \xrightarrow{Br_2/NaOH} CH_3CH_2CH_2NH_2$
   • Product is propylamine, formula $C_3H_9N$, a primary amine.
   • Gives positive carbylamine test. ✅

   B. $CH_3CON(CH_3)_2$

   • This is a tertiary amide (no $-CONH_2$ group).
   • Hofmann bromamide reaction requires a primary amide.
   • So this will not give the required primary amine. ❌

   C. $CH_3CH_2COCH_2NH$

   • This is not a proper primary amide structure for Hofmann degradation. ❌

   D. $CH_3COCH_2NHCH_3$

   • Also not a primary amide. ❌

5. Therefore, the only suitable compound is: $\boxed{CH_3CH_2CH_2CONH_2}$

So, option A is correct.