Step 1: Pick out what the quantity depends on:
We need the angle between two fringes, as seen from the slit. In a single slit of width a, the minima occur at $\sin\theta_n = n\lambda/a$.
Neither the slit-to-screen distance nor the screen position enters this relation.
Step 2: Test by doubling D:
Take two fringes at positions $y_1$ and $y_2$ on a screen at distance D. Then $\theta_1 \approx y_1/D$ and $\theta_2 \approx y_2/D$.
On moving the screen to $2D$, the fringe positions become $2y_1$ and $2y_2$, because each fringe stays on the same ray from the slit.
New angles: $2y_1/2D = y_1/D$ and $2y_2/2D = y_2/D$. They are the same as before.
Step 3: Check the options one by one:
Doubled (A): this describes the linear width, not the angular one.
Halved (B): angles would shrink only if the rays changed direction, and they do not.
One-fourth (D): there is no $1/D^2$ dependence here (that appears in intensity, not in angle).
Constant (C): matches the result $\theta = \lambda/a$.
Final Answer:
The angular separation of fringes stays the same, option (C).
\[ \boxed{\text{remains constant}} \]