I thought I would try
to derive your result.
After a number of mistakes,
here it is.
I agree - a simpler proof would be nice.
Using
$\cos(x)
=\frac12(e^{ix}+e^{-ix})
$,
$\begin{array}\\
s_p(x)
&=\sum_{k=0}^{2p-1}\left( \cos\left(x+\frac{k \pi}{2p}\right)\right)^{2p}\\
&=\dfrac1{2^{2p}}\sum_{k=0}^{2p-1}\left( e^{i\left(x+\frac{k \pi}{2p}\right)}+e^{-i\left(x+\frac{k \pi}{2p}\right)}\right)^{2p}\\
&=\dfrac1{4^{p}}\sum_{k=0}^{2p-1}\sum_{j=0}^{2p}\binom{2p}{j}\left( e^{ij\left(x+\frac{k \pi}{2p}\right)}e^{-i(2p-j)\left(x+\frac{k \pi}{2p}\right)}\right)\\
&=\dfrac1{4^{p}}\sum_{k=0}^{2p-1}\sum_{j=0}^{2p}\binom{2p}{j}\left( e^{ij\left(x+\frac{k \pi}{2p}\right)-i(2p-j)\left(x+\frac{k \pi}{2p}\right)}\right)\\
&=\dfrac1{4^{p}}\sum_{k=0}^{2p-1}\sum_{j=0}^{2p}\binom{2p}{j}\left( e^{2ij\left(x+\frac{k \pi}{2p}\right)-i2p\left(x+\frac{k \pi}{2p}\right)}\right)\\
&=\dfrac1{4^{p}}e^{-2ipx}\sum_{k=0}^{2p-1}\sum_{j=0}^{2p}\binom{2p}{j}\left( e^{2ij\left(x+\frac{k \pi}{2p}\right)-ik\pi}\right)\\
&=\dfrac1{4^{p}}e^{-2ipx}\sum_{j=0}^{2p}\sum_{k=0}^{2p-1}\binom{2p}{j}\left( e^{2ij\left(x+\frac{k \pi}{2p}\right)-ik\pi}\right)\\
&=\dfrac1{4^{p}}e^{-2ipx}\sum_{j=0}^{2p}e^{2ijx}\binom{2p}{j}\sum_{k=0}^{2p-1}\left( e^{2ij\left(\frac{k \pi}{2p}\right)-ik\pi}\right)\\
&=\dfrac1{4^{p}}e^{-2ipx}\sum_{j=0}^{2p}e^{2ijx}\binom{2p}{j}\sum_{k=0}^{2p-1}\left( e^{\left(\frac{ijk \pi}{p}\right)-ik\pi}\right)\\
&=\dfrac1{4^{p}}e^{-2ipx}\sum_{j=0}^{2p}e^{2ijx}\binom{2p}{j}\sum_{k=0}^{2p-1}\left( e^{k\pi i\left(\frac{j}{p}-1\right)}\right)\\
&=\dfrac1{4^{p}}e^{-2ipx}\left(2pe^{2ipx}\binom{2p}{p}+\sum_{j=0,j\ne p}^{2p}e^{2ijx}\binom{2p}{j}\dfrac{1-e^{2p\pi i\left(\frac{j}{p}-1\right)}}{1-e^{\pi i\left(\frac{j}{p}-2\right)}}\right)\\
&=\dfrac{1}{4^{p}}e^{-2ipx}\left(2pe^{2ipx}\binom{2p}{p}+\sum_{j=0,j\ne p}^{2p}e^{2ijx}\binom{2p}{j}\dfrac{1-e^{2\pi i\left(j-p\right)}}{1-e^{\pi i\left(\frac{j}{p}-2\right)}}\right)\\
&=\dfrac{2p}{4^p}\binom{2p}{p}\\
\end{array}
$