Answer
$pH = 8.62$ in both cases
Work Step by Step
$n(NH_4NO_3)=5.15\ g\div80.04\ g/mol=0.064\ mol$
$n(NH_3)=0.10\ L\times 0.15\ M=0.015\ mol$
Since $[NH_4NO_3]=n(NH_4NO_3)/V$, $[NH_3]=n(NH_3)/V$
$[NH_3]/[NH_4NO_3]=n(NH_3)/n(NH_4NO_3)$, i.e it's independent of the volume so diluting the solution won't change the pH of the buffer.
By the Henderson-Hasselbalch equation:
$pH=pKa+\log([NH_3]/[NH_4NO_3])$
$pKb(NH_3)=4.75 \rightarrow pKa = 14-4.75=9.25$
$pH=9.25+\log(0.015/0.064)$
$pH=9.25-0.63$
$pH = 8.62$
