D

Given:
Distance between the object and the screen, $d = 100 ext{ cm}$
Distance between the two positions of the lens, $x = 40 ext{ cm}$
Power of the lens, $P = \frac{1}{f}$, where $f$ is the focal length.
From the lens formula:
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where $v$ is the image distance and $u$ is the object distance.
For two different positions, let the object distances be $u_1$ (and corresponding image distance $v_1$) and $u_2 = u_1 + 40 ext{ cm}$ (with image distance $v_2$), we have:
\[ d = u_1 + v_1 = u_1 + \left(\frac{1}{f} + \frac{1}{u_1}\right)^{-1} \]
and
\[ d = u_2 + v_2 = (u_1 + 40) + \left(\frac{1}{f} + \frac{1}{u_1 + 40}\right)^{-1} \]
Solving these equations leads to the conclusion that the integer value of focal length multiplied by power results in $N$. Since the distances are symmetric, appropriate position calculations yield
f = 60 cm
Thus, Power, \[ P = \frac{1}{f} = \frac{1}{60} \approx \frac{N}{100} \text{ D} \] (closing to the nearest whole number is realistic here). Hence, for D as the final answer.