We have,
$$
f(x)=\left\{\begin{array}{cc}
\frac{1-\sqrt{2} \sin x}{\pi-4 x}, & x \neq \frac{\pi}{4} \\
k & , \quad x=\frac{\pi}{4}
\end{array}\right.
$$
Since, $f(x)$ is continuous at
$$
\begin{aligned}
& x=\frac{\pi}{4} \\
\therefore & f(\pi / 4)=\lim _{\pi \rightarrow \frac{\pi}{4}} f(x) \\
\Rightarrow \quad k & =\lim _{\pi \rightarrow \frac{\pi}{4}} \frac{1-\sqrt{2} \sin x}{\pi-4 x} \\
\Rightarrow \quad k & =\lim _{\pi \rightarrow \frac{\pi}{4}} \frac{-\sqrt{2} \cos x}{-4} \text { [using } L^{\prime} \text { hospital Rule] } \\
\Rightarrow \quad k & =\frac{\sqrt{2}}{4} \times \frac{1}{\sqrt{2}} \\
\Rightarrow \quad & k=\frac{1}{4} .
\end{aligned}
$$