Edition of 28.07.23
Let $C=\cos\dfrac k n\pi,\quad S=\sin\dfrac k n\pi,\quad T=\tan\dfrac k n\pi,\quad$ then
$$Q_k(p,q)=\csc\left(p+\frac k n\pi\right)\csc\left(q+\frac k n\pi\right)
=\dfrac1{\left(C\sin p+S\cos p\right)\left(C\sin q+S\cos q\right)}$$
$$=\dfrac1{C^2\cos p\cos q(T+\tan p)(T+\tan q)},$$
$$Q_k(p,q)=\dfrac{(1+T^2)\sec p \sec q}{
\left(T+\tan p\right)\left(T+\tan q\right)},\tag1$$
$$Q_k(p,q)=\sec p \sec q\left(1+\dfrac E{T+\tan p}+\dfrac F{T+\tan q}\right),$$
where
$$E=\lim_{T\to -\tan p}(T+\tan p)Q_k(p,q)\cos p\cos q=\dfrac{1+\tan^2 p}{\tan q-\tan p},$$
$$F=\lim_{T\to -\tan q}(T+\tan q)Q_k(p,q)\cos p\cos q =\dfrac{1+\tan^2 q}{\tan p-\tan q}.$$
Therefore,
$$Q_k(p,q)=\dfrac{\sec p\sec q}{\tan q-\tan p}\left(\tan q-\tan p+\dfrac{1+\tan^2 p}{T+\tan p}-\dfrac {1+\tan^2 q}{T+\tan q}\right).$$
$$Q_k(p,q)=\dfrac{h(q)-h(p)}{\sin(q-p)},\tag2$$
where
$$h(p)=\tan p-\dfrac{1+\tan^2 p}{T+\tan p}
=\dfrac{T\tan p-1}{T+\tan p},$$
$$\dfrac{h(p)h(q)-\tan p\tan q}{\tan p \tan q+1}
=\dfrac1{\tan p \tan q+1}\left(\dfrac{T\tan p-1}{T+\tan p}\,\dfrac{T\tan q-1}{T+\tan q} - \tan p \tan q\right)$$
$$=\dfrac1{\tan p \tan q+1}\left(\dfrac{T\tan p-1}{T+\tan p}\,\dfrac{T\tan q-1}{T+\tan q}+1 - (\tan p \tan q+1)\right)$$
$$=\dfrac{(T\tan p-1)(T\tan q-1)+(T+\tan p)(T+\tan q)}{(\tan p \tan q+1)(T+\tan p)(T+\tan q)}-1$$
$$=\dfrac{T^2(1+\tan p \tan q)+(1+\tan p\tan q)}{(\tan p \tan q+1)(T+\tan p)(T+\tan q)}-1
= \dfrac{1+T^2}{(T+\tan p)(T+\tan q)}-1$$
$$=Q_k(p,q)\cos p\cos q-1,$$
i.e.
$$h(p)h(q)=(Q_k(p,q)\cos p\cos q-1)(1+\tan p\tan q)+\tan p \tan q,$$
$$h(p)h(q)=Q_k(p,q)\cos(p-q)-1,\tag3$$
From OP,
$$\sum\limits_{k=0}^{n-1} Q_k(p,q) = \dfrac{n\sin n(p-q)}{\sin(p-q)\sin np\sin nq}.\tag4$$
Then,
$$\sum\limits_{k=0}^{n-1} h(p)h(q) = \dfrac{n\sin n(p-q)\cot(p-q)}{\sin np\sin nq}-1.\tag5$$
Therefore,
$$\begin{align}
&\sigma =\sum\limits_{k=0}^{n-1} \csc\left(w+\frac k n\pi\right)\csc\left(x+\frac k n\pi\right) \csc\left(y+\frac k n\pi\right)\csc\left(z+\frac k n\pi\right)\\[4pt]
&=\sum\limits_{k=0}^{n-1}Q_k(w,x) Q_k(y,z)
=\dfrac1{\sin(x-w)\sin(z-y)}\sum\limits_{k=0}^{n-1}(h(x)-h(w))(h(z)-h(y))\\[4pt]
&=\dfrac1{\sin(x-w)\sin(z-y)}\left(\sum\limits_{k=0}^{n-1} h(x)h(z)-\sum\limits_{k=0}^{n-1} h(x)h(y)-\sum\limits_{k=0}^{n-1} h(w)h(z)+\sum\limits_{k=0}^{n-1} h(w)h(y)\right),
\end{align}$$
$$\color{green}{\mathbf{\begin{align}
&\sigma=\dfrac{n}{\sin(x-w)\sin(z-y)}
\left(\dfrac{\sin n(x-z)\cot(x-z)}{\sin nx\sin nz}
-\dfrac{\sin n(x-y)\cot(x-y)}{\sin nx\sin ny}\right.\\[4pt]
&\left.-\dfrac{\sin n(w-z)\cot(w-z)}{\sin nw\sin nz}
+\dfrac{\sin n(w-y)\cot(w-y)}{\sin nw\sin ny}\right).
\end{align}}}$$
Old Edition
$$\begin{align}
&P_k=\csc\left(w+\frac k n\pi\right)\csc\left(x+\frac k n\pi\right) \csc\left(y+\frac k n\pi\right)\csc\left(z+\frac k n\pi\right)\\[4pt]
&=\dfrac1{\left(C\sin w+S\cos w\right)\left(C\sin x+S\cos x\right)
\left(C\sin y+S\cos y\right)\left(C\sin z+S\cos z\right)},\\[4pt]
&=\dfrac{\sec w\sec x\sec y\sec z\,(1+T^2)^2}{(T+\tan w)(T+\tan x)(T+\tan y)(T+\tan z)},\\[4pt]
&=\sec w\sec x\sec y\sec z\left(1+\dfrac A{T+\tan w}+\dfrac B{T+\tan x}+\dfrac C{T+\tan y}+\dfrac D{T+\tan z}\right),\\[4pt]
\end{align}$$
where
$$\begin{align}
&A=\cos w\cos x\cos y\cos z\lim\limits_{T\to -\tan w} (T+\tan w)P_k\\[4pt]
&=\dfrac{\cos x\cos y\cos z}{\cos^3w(\tan x-\tan w)(\tan y-\tan w)(\tan z-\tan w)},\\[4pt]
&B=\cos w\cos x\cos y\cos z\lim\limits_{T\to -\tan x} (T+\tan x)P_k\\[4pt]
&=\dfrac{\cos w\cos y\cos z}{\cos^3x(\tan w-\tan x)(\tan y-\tan x)(\tan z-\tan x))},\\[4pt]
&C=\cos w\cos x\cos y\cos z\lim\limits_{T\to -\tan y} (T+\tan y)P_k\\[4pt]
&=\dfrac{\cos w\cos x\cos z}{\cos^3y(\tan w-\tan y)(\tan x-\tan y)(\tan z-\tan y))},\\[4pt]
&D=\cos w\cos x\cos y\cos z\lim\limits_{T\to -\tan z} (T+\tan z)P_k\\[4pt]
&=\dfrac{\cos w\cos x\cos y}{\cos^3z(\tan w-\tan z)(\tan x-\tan z)(\tan y-\tan z))},\\[4pt]
\end{align}$$
My first idea was to use $\;Q_k(w,x),\; Q_k(x,y)\;$ etc. for definition of the terms $\;\dfrac1{T+\tan w},\; \dfrac1{T+\tan x},\; \dfrac1{T+\tan y},\; \dfrac1{T+\tan z}.\;$
But the rank of the obtained system is insufficient for this task.
Current idea is to express $\;P_k\;$ via the set of $\;Q_k.$ But I did not have success yet.