moles of $HCl = 0.2\,M \times 25 \times {10^{ - 3}}\,L = 0.005$

moles $HCl$ (total consumed)

moles of $N{H_3} = 0.2\,M \times 50 \times {10^{ - 3}}\,L = 0.01$

moles $HCl$

excess $N{H_3} = 0.01 - 0.005 = 0.005$ moles

$1$ mole ammonia $= 1 $ mole $N{H_4}Cl$

$0.005\,N{H_3} = 0.005\,N{H_4}Cl$

Total volume 

$ = {V_{HCl}} + {V_{N{H_3}}} = 25 + 50\, = 75\,ml$

$[N{H_3}] = [N{H_4}Cl] = \frac{{0.005\,mole}}{{75 \times {{10}^{ - 3}}\,L}} = 0.066\,M$

$pOH = p{K_b} + \log \frac{{[N{H_4}Cl]}}{{[N{H_3}]}}$

$pOH = 4.75 + \log \frac{{[0.066]}}{{[0.066]}}$

$pOH = 4.75$

$pH = 14 - pOH \Rightarrow pH = 9.25$