Let \[P(x) = 24x^{24} + \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).\] Let $z_{1},z_{2},\ldots,z_{r}$ be the distinct zeros of $P(x),$ and let $z_{k}^{2} = a_{k} + b_{k}i$ for $k = 1,2,\ldots,r,$ where $a_{k}$ and $b_{k}$ are real numbers. Let

$\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},$
where $m, n,$ and $p$ are integers and $p$ is not divisible by the square of any prime. Find $m + n + p.$
This can be factored as:
\[P(x) = x\left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right)^2\]
Note that $\left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right) \cdot (x-1) = x^{24} - 1$.
So the roots of $x^{23} + x^{22} + \cdots + x^2 + x + 1$ are exactly all $24$-th complex roots of $1$, except for the root $x=1$. 
Let $\omega=\cos \frac{360^\circ}{24} + i\sin \frac{360^\circ}{24}$. Then the distinct zeros of $P$ are $0,\omega,\omega^2,\dots,\omega^{23}$.
We can clearly ignore the root $x=0$ as it does not contribute to the value that we need to compute. 
The squares of the other roots are $\omega^2,~\omega^4,~\dots,~\omega^{24}=1,~\omega^{26}=\omega^2,~\dots,~\omega^{46}=\omega^{22}$.
Hence we need to compute the following sum:
\[R = \sum_{k = 1}^{23} \left|\, \sin \left( k\cdot \frac{360^\circ}{12} \right) \right|\]
Using basic properties of the sine function, we can simplify this to
\[R = 4 \cdot \sum_{k = 1}^{5} \sin \left( k\cdot \frac{360^\circ}{12} \right)\]
The five-element sum is just $\sin 30^\circ + \sin 60^\circ + \sin 90^\circ + \sin 120^\circ + \sin 150^\circ$. 
We know that $\sin 30^\circ = \sin 150^\circ = \frac 12$, $\sin 60^\circ = \sin 120^\circ = \frac{\sqrt 3}2$, and $\sin 90^\circ = 1$.
Hence our sum evaluates to:
\[R = 4 \cdot \left( 2\cdot \frac 12 + 2\cdot \frac{\sqrt 3}2 + 1 \right) = 8 + 4\sqrt 3\]
Therefore the answer is $8+4+3 = \boxed{015}$.