Finding final pH of the buffer solution without applying Henderson equation
As the amounts of substance in the final solution are known to be $n(\ce{NH3}) = \pu{2 mmol},$ and $n(\ce{NH4^+}) = \pu{0.5 mmol},$ you may simply use the definition of the constant $K_\mathrm{b}:$
$$K_\mathrm{b} = \frac{n(\ce{NH4^+})[\ce{OH^-}]}{n(\ce{NH3})} = \frac{\pu{0.5 mmol}\times [\ce{OH-}]}{\pu{2 mmol}} = \pu{3.3E-5}$$
from where $[\ce{OH-}],$ $[\ce{H+}]$ and $\mathrm{pH}$ can be quickly obtained:
$$[\ce{OH-}] = \pu{1.32E-4 mol L^-1}$$
$$[\ce{H+}] = \frac{10^{-14}}{[\ce{OH-}]} = \pu{7.57E-11 mol L^-1}$$
$$\mathrm{pH} = -\log[\ce{H+}] = 10.12$$