1.202 g of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt (ACES–K , MW = 220.29 g/mol) is dissolved in 90.00 mL of water...?

Let y M be the molarity of the HCl solution.

No. of moles of ACES⁻ added = No. of moles of ACES–K added = (1.202 g) / (220.29 g/mol) = 0.005456 mol
No. of moles of HCl added = (y mol/L) × (15.19/1000 L) = 0.01519y mol

HCl completely reacts with a part of ACES-K to form
ACES-K + HCl → ACES + KCl
After reaction :
No. of moles of ACES formed = No. of moles HCl added = 0.01519y mol
No. of moles of ACES⁻ unreacted left in the solution = (0.005456 - 0.01519y) mol
In the final solution, [ACES⁻] / [ACES] = 0.01519y / (0.005456 - 0.01519y)

Consider the dissociation of ACES :
ACES ⇌ ACES⁻ + H⁺

pH = pKa + log{[ACES⁻]/[ACES]}
6.62 = 6.85 + log{0.01519y / (0.005456 - 0.01519y)}
log{0.01519y / (0.005456 - 0.01519y)} = -0.23
0.01519y / (0.005456 - 0.01519y) = 10^(-0.23)
0.01519y = 0.003213 - 0.008945y
0.02414y = 0.003213
y = 0.1331

Hence, molarity of the HCl solution = 0.1331 M

Moles ACES-K = 1.202 g / 220.29 g/mol = 5.456X10^-3 moles

Addition of HCl will quantitatively convert the salt into its conjugate acid.

Use the Henderson-Hasselbalch equation to calculate the ratio of [ACES-] / [HACES]

pH = pKa + log [ACES-] / [HACES]
6.62 = 6.85 + [ACES-] / [HACES]
[ACE-S] / [HACES] = 0.589

Rearranging gives:
[ACES-] = 0.589 [HACES]
Because these are in the same volume of solution, we can deal with moles of each component rather than concentrations, so:
moles ACES- = 0.588 moles HACES

You also know that moles ACES- + moles HACES = 5.456X10^-3

Substituting the previous equation into this one gives:
0.589 mols HACES + mols HACES = 5.456X10^-3
moles HACES = 3.433X10^-3 moles

This must equal the moles of HCl added, so,
Molarity HCL = 3.433X10^-3 mol / 0.01519 L = 0.2260 M HCl