How do I do this question step by step?

First write the reaction of HCl (acid) and ammonia (base)
NH3 (aq) + HCl(aq)    NH4+ (aq) + Cl-(aq) 
Now you are mixing volume of two solutions so their concentrations are changing. Work with moles and millimoles to know how much of the reactants have reacted and how much of the product is formed.
Remember – molarity = mmol /V in mL  =  mol / Vol (L)     
Means mol = M* V in L
Here volume is not given so let us assume the volume is  1L
Reaction
              NH₃(aq)+HCl(aq)-> NH₄⁺(aq)+Cl^-(aq)

Initial (Mol      0.12          0.10              0                    0
Final (Mol))     0.02          0.00            0.10                   0.10
HCl is the limiting reactant and finally solution has NH₃ (aq), NH₄⁺ (aq)  and Cl^-(aq) 
Cl^-(aq)   is a spectator here because it is a weak conjugate base of strong acid HCl.
pH is controlled by NH₃ (aq) and  NH₄⁺ (aq)  .However this is a base and its conjugate acid so it is a buffer and you can use Henderson - Hasselbalch equation to get the answer.
pH = pKa + log ([ NH₃ ]/[NH₄⁺ ] )
Since volume is doubled you can calculate new concentrations. New volume is 2L
[ NH3 ]  = 0.02 /2 = 0.01 M
[NH4+ ]  = 0.10 / 2 = 0.05 M
pH = 9.25 + log ( 0.01/ 0.05)
     Or you can simply take the ratio of moles since volume finally is same in the buffer solution for both acid and its conjugate base.
 pH = pKa + log (moles  NH₃ /moles NH₄⁺  )
          = 9.25 + log ( 0.02/ 0.10)
log ( 0.01/ 0.05)  = log ( 0.02/ 0.10)