1. Key principle

A compound liberates $CO_2$ with $NaHCO_3$ only if it is more acidic than carbonic acid and can react as:

$HA + NaHCO_3 \rightarrow NaA + H_2CO_3$

and then

$H_2CO_3 \rightarrow CO_2 + H_2O$

So we need a compound that can provide a sufficiently acidic proton.

2. Check each option

Option A: $(CH_3)_3NH^+Cl^-$

This is trimethylammonium chloride, an ammonium salt.

The cation $(CH_3)_3NH^+$ can donate a proton:

$(CH_3)_3NH^+ + HCO_3^- \rightarrow (CH_3)_3N + H_2CO_3$

and then

$H_2CO_3 \rightarrow CO_2 + H_2O$

Hence, $CO_2$ is liberated.

Option B: $(CH_3)_4N^+OH^-$

This is tetramethylammonium hydroxide, a base, not an acid.

Since it does not provide an acidic proton, it will not liberate $CO_2$ with $NaHCO_3$.

Option C:

The option is not visible/provided in the question statement. So only the shown options can be evaluated.

Option D: $CH_3NH_2$

This is methylamine, a base.

It is not acidic enough to react with $NaHCO_3$ to produce $CO_2$.

So it will not liberate $CO_2$.

3. Conclusion

Among the given options, only Option A contains an acidic proton on the ammonium ion and reacts with $NaHCO_3$ to produce $CO_2$.

$\boxed{A}$

4. Comparison with stored answer

Stored correct answer: A

My derived answer: A

They match.