Identify the reaction involved

$p$-Toluenesulfonyl chloride reacts with amines in the Hinsberg test.

General behavior:

• Primary amines react to form sulfonamides having an acidic $\mathrm{N-H}$ proton, so they are soluble in aqueous NaOH.
• Secondary amines react to form sulfonamides with no acidic hydrogen on nitrogen, so they are insoluble in aqueous NaOH.
• Tertiary amines generally do not react with sulfonyl chloride because there is no $\mathrm{N-H}$ proton required for sulfonamide formation.

Examine Statement (I)

The compounds given are:

• $\mathrm{C_6H_5NH_2}$ = aniline = primary amine
• $\left(\mathrm{C_6H_5}\right)_2\mathrm{NH}$ = diphenylamine = secondary amine
• $\left(\mathrm{C_6H_5}\right)_3\mathrm{N}$ = triphenylamine = tertiary amine

Now check reactivity with $p$-toluenesulfonyl chloride:

• Aniline: reacts
• Diphenylamine: reacts
• Triphenylamine: does not react

Therefore, the statement "All the following compounds react with $p$-toluenesulfonyl chloride" is false.

So, Statement (I) is false.

Examine Statement (II)

Statement (II): "Their products in the above reaction are soluble in aqueous $\mathrm{NaOH}$."

Check product solubility:

• Product from aniline (primary amine) is a sulfonamide containing acidic $\mathrm{N-H}$, hence soluble in aqueous NaOH.
• Product from diphenylamine (secondary amine) is an $N,N$-disubstituted sulfonamide with no acidic hydrogen, hence insoluble in aqueous NaOH.
• Triphenylamine does not react, so no such product forms.

Hence it is incorrect to say that the products are soluble in aqueous $\mathrm{NaOH}$.

So, Statement (II) is also false.

Conclusion

• Statement (I): False
• Statement (II): False

Therefore, the correct option is: $\boxed{\text{C: Both Statement I and Statement II are false}}$

Comparison with stored answer

Stored correct answer = C.

My derived answer also = C. Hence, they agree.