The expression $\sum_{k=1}^{32}(3K + 2)\left\{\sum_{r=1}^{10}\left(\sin\dfrac{2r\pi}{11} - i\cos\dfrac{2r\pi}{11}\right)\right\}^{k}$ represents

Let $S = \sum_{r=1}^{10}\left(\sin\dfrac{2r\pi}{11} - i\cos\dfrac{2r\pi}{11}\right)$

$S = -i\sum_{r=1}^{10}\left(\cos\dfrac{2r\pi}{11} + i\sin\dfrac{2r\pi}{11}\right) = -i\sum_{r=1}^{10}e^{i\dfrac{2r\pi}{11}}$

Since the sum of all 11th roots of unity is zero, $1 + \sum_{r=1}^{10}e^{i\dfrac{2r\pi}{11}} = 0 \Rightarrow \sum_{r=1}^{10}e^{i\dfrac{2r\pi}{11}} = -1$

$S = -i(-1) = i$

The given expression becomes $P = \sum_{k=1}^{32}(3k + 2)i^{k}$

We can group the terms into 8 blocks of 4 terms each. For any integer $m \ge 0$, the sum of 4 consecutive terms starting from $k = 4m+1$ is:

$(3(4m+1)+2)i^{4m+1} + (3(4m+2)+2)i^{4m+2} + (3(4m+3)+2)i^{4m+3} + (3(4m+4)+2)i^{4m+4}$

$= (12m+5)i + (12m+8)i^2 + (12m+11)i^3 + (12m+14)i^4$

$= (12m+5)i - (12m+8) - (12m+11)i + (12m+14)$

$= 6 - 6i$

Since there are $32$ terms, there are exactly $8$ such blocks.

$P = 8(6 - 6i) = 48(1 - i)$

Answer: $48(1 - i)$