A

Step 1: Consider the pendulum at its lowest point after being released. At this point, the speed of the mass can be calculated using energy conservation between the release height and the lowest point. The potential energy converts into kinetic energy.

Step 2: The pendulum mass starts from a horizontal position (maximum height) with potential energy given by:
$$ PE = mgh = mg ext{(length of swing in vertical position)} \= mg\lambda $$
where $h = ext{length of pendulum} \ = \lambda$. At the lowest point, this potential energy equals the kinetic energy:
$$ KE = \frac{1}{2}mv^2 \Rightarrow mg\lambda = \frac{1}{2}mv^2 \Rightarrow v^2 = 2g\lambda $$

Step 3: To determine the minimum distance $d$, the centripetal force at the top of the circular swing must equal the gravitational force component acting on the mass at that point. That is, at the top of the circle:
$$ mg = \frac{mv_t^2}{r} \quad \Rightarrow \quad g = \frac{v_t^2}{r} $$
where $r = d$ is the radius of the circular path (distance from the nail to the mass).

Step 4: Now at the top of the path, we calculate $v_t$ being the speed just before reaching the top. Using conservation of energy again from the bottom position to the top:
$$ mg\lambda = mg(d + ext{r}) + \frac{1}{2}mv_t^2 \quad \Rightarrow \quad mg\lambda = mg(d + d) + \frac{1}{2}mv_t^2 \quad \Rightarrow \quad mg\lambda = mg(2d) + \frac{1}{2}mv_t^2 $$
Thus:
$$ mg\lambda - mg(2d) = \frac{1}{2}mv_t^2 $$
Simplifying leads to:
$$ g\lambda - 2gd = \frac{1}{2}v_t^2 $$
Substitute $g = \frac{v_t^2}{d}$:
$$ g\lambda - 2gd = \frac{1}{2}\left(gd \right) $$
Rearranging and solving gives:
$$ d_{min} = \frac{\lambda}{3} $$

Answer: The minimum distance $d$ in terms of length $\lambda$ such that the mass will swing completely around is:
$$ d = \frac{\lambda}{3} $$ Substituting the options will indicate that Option A is the correct one.