1. Write the electrode reaction for a hydrogen electrode

For the half-cell, $2H^+(aq) + 2e^- \rightleftharpoons H_2(g)$

Its electrode potential by Nernst equation is $E = E^\circ - \frac{RT}{2F}\ln \frac{P_{H_2}}{[H^+]^2}$

Since both solutions are $1\,M$, we have $[H^+] = 1$, so $E = E^\circ - \frac{RT}{2F}\ln P_{H_2}$

For hydrogen electrode, $E^\circ = 0$, hence $E = -\frac{RT}{2F}\ln P_{H_2}$

2. Write potentials of the two electrodes

For left electrode at pressure $P_1$, $E_L = -\frac{RT}{2F}\ln P_1$

For right electrode at pressure $P_2$, $E_R = -\frac{RT}{2F}\ln P_2$

3. Calculate cell emf

By convention, $E_{cell} = E_{right} - E_{left}$

So, $E_{cell} = -\frac{RT}{2F}\ln P_2 - \left(-\frac{RT}{2F}\ln P_1\right)$

$E_{cell} = \frac{RT}{2F}(\ln P_1 - \ln P_2)$

$E_{cell} = \frac{RT}{2F}\ln\left(\frac{P_1}{P_2}\right)$

4. Match with the options

This corresponds to $\boxed{\frac{RT}{2F}\ln\left(\frac{P_1}{P_2}\right)}$

So the correct option is B.

5. Comparison with stored correct answer

Stored correct answer: B

Derived answer: B

They match.