Let the figure have 3 rows with boxes distributed as shown in the standard pattern for this question:

• Top row: $2$ boxes
• Middle row: $3$ boxes
• Bottom row: $3$ boxes

So total boxes $=2+3+3=8$.

We must place the 5 distinct letters $A,B,C,D,E$ into these 8 boxes such that:

1. At most one letter is placed in a box.
2. No row remains empty.

1. Total unrestricted placements

Choose 5 of the 8 boxes and arrange the 5 distinct letters in them:

${}^8P_5 = \frac{8!}{3!}=6720$

This counts all placements, including those where one or more rows may be empty.

2. Subtract arrangements where some row is empty

We use inclusion-exclusion.

Let:

• $E_1$ = top row empty
• $E_2$ = middle row empty
• $E_3$ = bottom row empty

We count each.

Case 1: Top row empty

Then all 5 letters must be placed in the remaining $6$ boxes.

Number of ways:

${}^6P_5 = 6\cdot5\cdot4\cdot3\cdot2 = 720$

Case 2: Middle row empty

Then only top + bottom rows are available: $2+3=5$ boxes. All 5 letters must occupy all 5 boxes.

Number of ways:

$5! = 120$

Case 3: Bottom row empty

Similarly, available boxes = $2+3=5$.

Number of ways:

$5! = 120$

So total to subtract initially:

$720+120+120=960$

3. Check pairwise intersections

If two rows are empty, all 5 letters would have to be placed in the boxes of the remaining single row. But the rows have sizes $2,3,3$, none of which can hold 5 letters.

Hence,

$E_1\cap E_2 = E_2\cap E_3 = E_3\cap E_1 = \varnothing$

and triple intersection is also impossible.

So no further correction is needed.

4. Final count

Therefore, required number of ways:

$6720-960=5760$

5. Option check

• A: $5880$ ❌
• B: $840$ ❌
• C: $960$ ❌
• D: $5760$ ✅

So the correct answer is:

$\boxed{5760}$

Hence, Option D is correct.