The value of $\cos y \cos \left(\frac{\pi}{2} - x\right) - \cos \left(\frac{\pi}{2} - y\right) \cos x + \sin y \cos \left(\frac{\pi}{2} - x\right) + \cos x \sin \left(\frac{\pi}{2} - y\right)$ is zero if

Use $\cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta$ and $\sin\left(\frac{\pi}{2}-\theta\right)=\cos\theta$. Then the expression becomes $$\cos y\sin x-\sin y\cos x+\sin y\sin x+\cos x\cos y.$$ Group the terms: $$\left(\cos y\sin x+\sin y\sin x\right)+\left(\cos x\cos y-\sin y\cos x\right).$$ Factor: $$\sin x\left(\cos y+\sin y\right)+\cos x\left(\cos y-\sin y\right).$$ Now use $$\cos y+\sin y=\sqrt{2}\cos\left(y-\frac{\pi}{4}\right),$$ and $$\cos y-\sin y=\sqrt{2}\cos\left(y+\frac{\pi}{4}\right).$$ So the expression is zero when $$\sin x\left(\cos y+\sin y\right)+\cos x\left(\cos y-\sin y\right)=0.$$ Testing the options directly is easiest. For $x=y$, $$\cos y\sin y-\sin y\cos y+\sin^2 y+\cos^2 y=1,$$ so it is not zero. For $x=0$, $$0-sin y+0+\cos y,$$ which is not always zero. For $y=0$, $$\sin x+\cos x,$$ which is not always zero. For $x=n\pi+y-\frac{\pi}{4}$, re-reading the options and matching the derived condition gives the zero case, so the correct option is $\text{D}$.