The reduction potential of hydrogen half cell will be negative if:

For the hydrogen electrode the half-reaction is
$$2\,\text{H}^{+}(aq)+2e^- \rightarrow \text{H}_2(g)$$

The Nernst equation at $$298\text{ K}$$ is
$$E = E^{\circ} - \frac{0.0591}{n}\,\log Q$$
where $$n = 2$$ and the reaction quotient is
$$Q = \frac{p(\text{H}_2)}{[\text{H}^+]^{2}}$$

Because $$E^{\circ}=0\,$$ V for a standard hydrogen electrode, we get
$$E = -\frac{0.0591}{2}\,\log\!\left(\frac{p(\text{H}_2)}{[\text{H}^+]^{2}}\right) \qquad -(1)$$

The sign of $$E$$ depends only on the sign of the logarithm:
• If $$\log(\ldots) \gt 0$$, the quantity inside is >1 and the term subtracted in (1) is positive, making $$E$$ negative.
• If $$\log(\ldots) \lt 0$$, the ratio is <1 and $$E$$ becomes positive.

Evaluate each option:

Option A: $$p=1\,\text{atm},\;[\text{H}^+]=1\,\text{M}$$
$$\frac{p}{[\text{H}^+]^{2}} = \frac{1}{1^{2}} = 1\;\Longrightarrow\;\log 1 = 0$$
Hence $$E = 0$$ (neither positive nor negative).

Option B: $$p=2\,\text{atm},\;[\text{H}^+]=1\,\text{M}$$
$$\frac{p}{[\text{H}^+]^{2}} = \frac{2}{1^{2}} = 2\;\Longrightarrow\;\log 2 \gt 0$$
In (1) we subtract a positive number, so $$E$$ is negative.

Option C: $$p=2\,\text{atm},\;[\text{H}^+]=2\,\text{M}$$
$$\frac{p}{[\text{H}^+]^{2}} = \frac{2}{(2)^{2}} = 0.5\;\Longrightarrow\;\log 0.5 \lt 0$$
Subtracting a negative value makes $$E$$ positive.

Option D: $$p=1\,\text{atm},\;[\text{H}^+]=2\,\text{M}$$
$$\frac{p}{[\text{H}^+]^{2}} = \frac{1}{(2)^{2}} = 0.25\;\Longrightarrow\;\log 0.25 \lt 0$$
Again $$E$$ is positive.

Only Option B yields a positive logarithm and therefore a negative electrode potential.

Final answer: Option B which is: $$p(\text{H}_2)=2\,\text{atm}$$ and $$[\text{H}^+]=1.0\,\text{M}$$.