Let the A.P. be $a_1,\, a_2,\, a_3,\dots$ with first term $a_1$ and common difference $d$.

Given that the sum of the first $11$ terms is $0$: $S_{11} = \frac{11}{2}\left[2a_1 + 10d\right] = 0$ Since $11 \neq 0$, $2a_1 + 10d = 0$ $a_1 + 5d = 0$ $d = -\frac{a_1}{5}$

Now consider the sequence $a_1,\, a_3,\, a_5,\dots, a_{23}$ These are the odd-numbered terms of the original A.P.

Number of terms from $a_1$ to $a_{23}$ taking only odd indices: $1,3,5,\dots,23$ This is an A.P. in indices with common difference $2$. Number of terms: $n = \frac{23-1}{2}+1 = 12$

The odd-indexed terms also form an A.P.:

• First term $= a_1$
• Common difference $= 2d$

So their sum is $\frac{12}{2}\left[2a_1 + (12-1)(2d)\right]$ $= 6\left[2a_1 + 22d\right]$

Substitute $d = -\frac{a_1}{5}$: $6\left[2a_1 + 22\left(-\frac{a_1}{5}\right)\right]$ $= 6\left[2a_1 - \frac{22a_1}{5}\right]$ $= 6\left[\frac{10a_1 - 22a_1}{5}\right]$ $= 6\left(-\frac{12a_1}{5}\right)$ $= -\frac{72}{5}a_1$

Hence, $a_1 + a_3 + a_5 + \cdots + a_{23} = k a_1$ gives $k = -\frac{72}{5}$

Therefore, the correct option is $\boxed{\text{D}}$