As the number of moles in the final solution are known to be $n \ce{(NH3) = 2}$ mol, and $n\ce{(NH4^+) = 0.5}$ mol, you may simply use the definition of the constant $K_b$ which is : $$\ce{K_b = \frac{n(NH4^+)[OH^-]}{n(NH3)} =  \frac{0.5·[OH-]}{2} = 3.3·10^{-5}}$$ from where $\ce{[OH-]}$, $\ce{[H+]}$ and $p$H can be quickly obtained :$$\ce{[OH^-] = 1.32·10^{-4}}$$ $$\ce{[H^+] = \frac{10^{-14}}{[OH-]} = 7.57·10^{-11}}$$ $$p\ce{H = - log [H+] = 10.12}$$