The equation of a plane passing through a point $\left(x_1, y_1, z_1\right)$ and having direction ratios of the normal to the plane as $<a, b, c>$ is $a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0.$

Thus, the equation of the plane passing through $\left(0, -1, 0\right)$ is
$a\left(x-0\right)+b\left(y+1\right)+c\left(z-0\right)=0$

$\Rightarrow ax+by+cz+b=0$

$\because$ It passes through $\left(0, 0, 1\right)$

$\Rightarrow c=-b$

$\therefore$ Equation of plane is $ax+by-bz+b=0 \dots \left(i\right)$

$\because$ It makes angle $\frac{\pi }{4}$ with plane $y-z+5=0$

The angle $\theta$ between the planes $a_{1}x+b_{1}y+c_{1}z+d_1=0$ and $a_{2}x+b_{2}y+c_{2}z+d_2=0$ is defined as $\cos \theta =\frac{\left|a_1{a}_2+b_1{b}_2+c_1{c}_2\right|}{\sqrt{a_1^2+b_1^2+c_1^2}\times \sqrt{a_2^2+b_2^2+c_2^2}}$

$\Rightarrow \cos \frac{\pi }{4}=\frac{\left|0+b+b\right|}{\sqrt{a^2+b^2+b^2}\times \sqrt{2}}$

$\Rightarrow \frac{1}{\sqrt{2}}=\frac{\left|2b\right|}{\sqrt{a^2+2b^2}\times \sqrt{2}}$

Squaring both sides, we get

$\Rightarrow a^2+2b^2=4b^2$

$\Rightarrow a^2=2b^2$

$\Rightarrow a=\pm \sqrt{2}b$

$\therefore$ Equation of plane is $\pm \sqrt{2}bx+by-bz+b=0$

$\Rightarrow \pm \sqrt{2}x+y-z+1=0$

$\therefore$ Option $\left(\sqrt{2}, 1, 4\right)$ satisfies the given equation.