$$
\begin{aligned}
&\begin{aligned}
&-\cos (p \sin x)=\sin (p \cos x) \\
\Rightarrow & \cos (p \sin x)=\cos \left(\frac{\pi}{2}-p \cos x\right)
\end{aligned}\\
&2 sin x \pm\left(\frac{\pi}{2}-p \cos x\right) \quad(n \geqslant 0)\\
&\Rightarrow p \sin x+p \cos x=2 n \pi+\frac{\pi}{2}\\
&4\\
&p \sin x-p \cos x=2 n \pi-\frac{\pi}{2}\\
&\Rightarrow p \sqrt{2}\left(\sin \left(x+\frac{\pi}{4}\right)\right)=\frac{(4 n+1)}{2} \pi
\end{aligned}
$$
or $$\frac{\sqrt{2}\left(\sin \left(x-\frac{\pi}{4}\right)\right)}=\frac{(4 n-1) n}{2}$$
As $$-1 \leq \sin \left(x+\frac{\pi}{4}\right) \leqslant 1$$
$$\Rightarrow-p \sqrt{2} \leq p \sqrt{2} \sin \left(x+\frac{\pi}{4}\right) \leqslant p \sqrt{2}$$
$$\Rightarrow-p \sqrt{2} \leq\left(\frac{4 n+1}{2}\right) \pi \leq p \sqrt{2} \quad$$... (i)
And similarly,
$$\Rightarrow-p \sqrt{2} \leq\left(\frac{4 n-1}{2}\right)^{n} \leq p \sqrt{2}$$
(ii) is subset of (i),
$$\therefore$$ We can consider only (i).
If $$\begin{array}{ll}n \geqslant 0, & \left(\frac{4 n+1}{2}\right) \pi \Rightarrow\left(\frac{4 n+1}{2}\right) n \leqslant \sqrt{2} p\end{array}$$
$$\Rightarrow \sqrt{2} p \geqslant \frac{\pi}{2}$$
$$\Rightarrow p \geqslant \frac{\pi}{2 \sqrt{2}} \Rightarrow \quad p \geqslant \frac{\pi \sqrt{2}}{4}$$
$$\therefore$$ option $$D$$ is correct.