The value of $a$ so that $f(x)=\sin x-\cos x-ax+b$ decreases for all real values of $x$ is given by:

For $f(x)$ to be a decreasing function for all $x$, $f^{\prime }(x)\le 0$ for all $x$.$f(x)=\sin x-\cos x-ax+b$$f^{\prime }(x)=\cos x+\sin x-a$ We need $\cos x+\sin x-a\le 0⟹a\ge \cos x+\sin x$.The expression $\cos x+\sin x$ can be rewritten as:$\sqrt{2}\left[\frac{1}{\sqrt{2}}\cos x+\frac{1}{\sqrt{2}}\sin x\right]=\sqrt{2}\sin (x+\frac{\pi }{4})$ The maximum value of $\sqrt{2}\sin (x+\frac{\pi }{4})$ is $\sqrt{2}$.For the inequality to hold for all $x$, $a$ must be greater than or equal to the maximum value of the expression.Therefore, $a\ge \sqrt{2}$.