The answer for this question is C but why would we use the Henderson Hasselbachs equation if it is not a buffer

Definitely it is not a buffer solution as you are dealing with strong acid and strong base .
But C is not the Handerson equation
It is based on pH + pOH = 14 equation written in a different format.
Write reaction first and as you are mixing volume of two solutions so their concentrations are changing. Work with moles or  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

Reaction
               NaOH(aq) +  HCl(aq) ->   Na+ (aq)  + Cl-(aq)  + H2O (l)

Initial (mMol)   11.0     10.0           0            0             -
Final (mMol))   1.0     0.00           10             10             -
HCl is the limiting reactant and finally solution has NaOH  (aq) left which is very strong base and it will control the pH .
Cl-(aq) and Na+ (aq) is a spectator here because they are weak conjugate base/acid  of strong acid and strong base. Water is also weak in front of NaOH.
Since volume has changed  you can calculate new concentrations. New volume is 21 mL = 0.021L
[ NaOH  ]  = mmol  /V mL  = [OH-]= 1.0 mmol/21mL 
Which can be written in terms of moles
[ NaOH  ]  = mol  /V L  = [OH-]= 0.0010 mol/0.021L 

pOH = -log[OH-] = -log (0.0010/0.021)
pH = 14- pOH
pH = 14 – (-log[OH-])
pH = 14+ log[OH-]
=14+ log (0.0010/0.021)
That's how C is the answer.