so I got your mistake you are not considering here that initially the ph of the solution is 10 .
so if we consider this then
Before adding hcl solution 10 = 9.25 + log [NH3]/[NH4+]
[NH3]/[NH4+] = 5.6234 -(1)
Also moles of (NH3 + NH4+) = 1M x 75 ml = 75 mmoles -(2)
=> on solving equation 1 and equation2 we get
moles of NH3 were 63.676533 mmoles
moles of NH4+ were 11.323466 mmoles
after adding the HCL solution , 45 mmoles of H+ will react with NH3 to produce more NH4
according to equation
NH3 + H+ ————> NH4 moles of NH3 will react with H+ to produce more NH4+
hence final moles of NH3 will be 63.676533 – 45 = 18.676533 mmoles
also final moles of NH4+ will be 11.323466 + 45 = 56.3234 mmoles
now you should apply the equation which you have applied in your solution
ph = 9.25 + log [18.676533]/[56.3234] = 9.25 – 0.48 = 8.77 .