Let's analyze both statements one by one.

Statement (I): All the following compounds react with p-toluenesulfonyl chloride.

The compounds given are:

$$\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 \quad (\mathrm{Aniline})$$

$$\left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \mathrm{NH} \quad (\mathrm{Diphenylamine})$$

$$\left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \mathrm{N} \quad (\mathrm{Triphenylamine})$$

P-toluenesulfonyl chloride (TsCl) reacts with amines to form sulfonamides. The general reaction is:

$$\mathrm{RNH}_2 + \mathrm{TsCl} \rightarrow \mathrm{RNHTs} + \mathrm{HCl}$$

For this reaction to occur, the amine must have at least one hydrogen attached to the nitrogen. Let's check the provided compounds:

1. $$\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$$ (Aniline) - primary amine, has 2 hydrogens, so it reacts.

2. $$\left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \mathrm{NH}$$ (Diphenylamine) - secondary amine, has 1 hydrogen, so it reacts.

3. $$\left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \mathrm{N}$$ (Triphenylamine) - tertiary amine, has no hydrogen, so it does not react.

Hence, Statement I is false, since $$\left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \mathrm{N}$$ does not react with p-toluenesulfonyl chloride.

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

The products of the reaction of aniline and diphenylamine with p-toluenesulfonyl chloride (if it were to happen for the sake of analysis) would be sulfonamides:

$$\mathrm{RNH}_2 + \mathrm{TsCl} \rightarrow \mathrm{RNHTs}$$

Sulfonamides (RNHTs) are generally not soluble in aqueous NaOH because they are neutral compounds and NaOH is a strong base. Only aniline itself is soluble in aqueous NaOH due to its basic nature, but its product with TsCl is not.

Thus, Statement II is also false, as the products would not be soluble in aqueous NaOH.

Conclusion:

Both Statement I and Statement II are false. Therefore, the correct answer is:

Option C

Both Statement I and Statement II are false.