1. Write dimensions of each quantity

We compare the dimensions option by option.

2. Option A: Electric displacement $\vec D$ and surface charge density

Using $\vec D = \varepsilon \vec E$

Also, from Gauss's law in matter, $\oint \vec D \cdot d\vec S = q_{\text{free}}$ So, $[D] = \frac{\text{charge}}{\text{area}}$

Dimension of charge: $[Q] = [IT]$ Hence, $[D] = \frac{[IT]}{[L^2]} = [I T L^{-2}]$

Now surface charge density $\sigma = \frac{Q}{A}$ so $[\sigma] = \frac{[IT]}{[L^2]} = [I T L^{-2}]$

Thus, $\vec D$ and surface charge density have the same dimensions.

So, Option A is correct.

3. Option B: Displacement current and electric field

Displacement current has dimensions of current: $[I_d] = [I]$

Electric field: $E = \frac{F}{q}$ So, $[E] = \frac{[M L T^{-2}]}{[I T]} = [M L T^{-3} I^{-1}]$

Clearly, $[I] \neq [M L T^{-3} I^{-1}]$

So, Option B is incorrect.

4. Option C: Current density and surface charge density

Current density: $J = \frac{I}{A}$ Hence, $[J] = [I L^{-2}]$

Surface charge density: $[\sigma] = [I T L^{-2}]$

These differ by a factor of $T$.

So, Option C is incorrect.

5. Option D: Electric potential and energy

Electric potential: $V = \frac{\text{work}}{\text{charge}}$ So, $[V] = \frac{[M L^2 T^{-2}]}{[I T]} = [M L^2 T^{-3} I^{-1}]$

Energy: $[U] = [M L^2 T^{-2}]$

These are not the same.

So, Option D is incorrect.

6. Final conclusion

Only Option A has the same dimensions in both quantities.