The line $x\cos \alpha +y\sin \alpha =p$ is a tangent to 

${\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n=2$ at $\left(a,b\right)$.

Differentiating ${\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n=2$, we get

$\frac{n}{a^n}{\left(x\right)}^{n-1}+\frac{n}{b^n}{\left(y\right)}^{n-1}\frac{dy}{dx}=0$

i.e. ${\left(\frac{dy}{dx}\right)}_{\left(a,b\right)}=-\frac{b}{a}$

Now equation of tangent at $\left(a,b\right)$ will be

$y-b=-\frac{b}{a}\left(x-a\right)\Rightarrow ay-ab=-bx+ab$

$\Rightarrow bx+ay=2ab$

Comparing with $x\cos \alpha +y\sin \alpha =p$, we get

$\cos \alpha =b, \sin \alpha =a, p=2ab$

i.e. $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{{\displaystyle {\cos }^2}{\displaystyle \alpha }}+\frac{1}{{\displaystyle {\sin }^2}{\displaystyle \alpha }}=\frac{1}{{\sin }^2\alpha {\cos }^2\alpha }$

Also, $\frac{k}{p^2}=\frac{k}{4a^2{b}^2}=\frac{k}{4{\sin }^2\alpha {\cos }^2\alpha }$

Hence, $k=4$