1. Identify the reaction

The given reaction is:

$CH_3CH_2NH_2 + CHCl_3 + 3KOH \rightarrow (A) + (B) + 3H_2O$

Here, $CH_3CH_2NH_2$ is a primary amine (ethylamine), and it is reacting with chloroform ($CHCl_3$) and alkali ($KOH$).

This is the carbylamine reaction (also called isocyanide test).

2. General form of carbylamine reaction

A primary amine reacts with chloroform and alcoholic KOH to give an isocyanide:

$RNH_2 + CHCl_3 + 3KOH \rightarrow RNC + 3KCl + 3H_2O$

So:

• the amine $RNH_2$ gives isocyanide $RNC$
• the inorganic salt formed is $3KCl$

3. Apply to the given amine

Here,

$R = C_2H_5$

Therefore,

$CH_3CH_2NH_2 \rightarrow C_2H_5NC$

So,

• $(A) = C_2H_5NC$
• $(B) = 3KCl$

4. Check the options

• Option A: $C_2H_5CN$ and $3KCl$
  Wrong, because $C_2H_5CN$ is a nitrile, not isocyanide.

• Option B: $CH_3CH_2CONH_2$ and $3KCl$
  Wrong, this is an amide and not formed in carbylamine reaction.

• Option C: $C_2H_5NC$ and $K_2CO_3$
  Wrong, salt formed is $KCl$, not $K_2CO_3$.

• Option D: $C_2H_5NC$ and $3KCl$
  Correct.

5. Final answer

$(A) = C_2H_5NC, \qquad (B) = 3KCl$

So the correct option is D.