By Nernst equation
$E_{H^+/H_2}=E^o_{H^+/H_2}-\large\frac{0.0591}{1}$$\log\large\frac{ \sqrt{P_{H_2}}}{[H^+]}$
Since cell reaction is
$H^+e^-\rightarrow \large\frac{1}{2}$$H_2(g)$
$P_{H_2}=1$atm
$E^o_{H^+/H_2}=0$
Standard hydrogen electrode as reference electrode
$E_{H^+/H_2}=-0.059\log \large\frac{1}{[H^+]}$
$E_{H^+/H_2}=-0.059\times P_H$
If we increase $[H^+]$ then $P_H$ increases to $E_{H^+/H_2}$ increases
Hence (b) is the correct answer.