The solubility of AgCl will be maximum in deionized water.

Silver chloride (AgCl) is a sparingly soluble salt. Its solubility in water can be represented by the equilibrium:

$$ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) $$

In the presence of a common ion, the solubility of AgCl decreases due to the common ion effect, which is explained by Le Chatelier's Principle. Here's an analysis of each option:

Option A: 0.01 M KCl

The presence of additional chloride ions ($\text{Cl}^-$) from KCl will shift the equilibrium to the left, reducing the solubility of AgCl.

Option B: 0.01 M HCl

Similar to KCl, HCl also provides more chloride ions, which will decrease the solubility of AgCl due to the common ion effect.

Option C: 0.01 M AgNO$_3$

AgNO$_3$ introduces additional silver ions ($\text{Ag}^+$) into the solution, also shifting the equilibrium to the left, hence decreasing the solubility of AgCl.

Option D: Deionized water

This option does not introduce additional ions that impact the equilibrium, so there are no additional silver or chloride ions to shift the equilibrium. As a result, the solubility of AgCl will be highest in deionized water, as there is no common ion effect.

Therefore, the solubility of AgCl is highest in deionized water.