Step 1: Set up the tangent at the given point.
The curve is $\left(\frac{x}{a}\right)^n+\left(\frac{y}{b}\right)^n=2$, and we examine the point $(a,b)$ which clearly lies on it since $1+1=2$.
Step 2: Differentiate to get the slope.
Differentiating, $\frac{n}{a}\left(\frac{x}{a}\right)^{n-1}+\frac{n}{b}\left(\frac{y}{b}\right)^{n-1}\frac{dy}{dx}=0$. At $(a,b)$ both bracket factors are $1$, so $\frac{n}{a}+\frac{n}{b}\frac{dy}{dx}=0$, giving slope $\frac{dy}{dx}=-\frac{b}{a}$.
Step 3: Write the tangent line.
The tangent at $(a,b)$ is $y-b=-\frac{b}{a}(x-a)$, i.e. $\frac{x}{a}+\frac{y}{b}=2$, or $bx+ay=2ab$.
Step 4: Match with the normal form.
The given touching line is $x\cos\alpha+y\sin\alpha=p$. Comparing coefficients with $bx+ay=2ab$, the ratios give $\frac{\cos\alpha}{b}=\frac{\sin\alpha}{a}=\frac{p}{2ab}$.
Step 5: Express $\cos\alpha$ and $\sin\alpha$.
Thus $\cos\alpha=\frac{pb}{2ab}=\frac{p}{2a}$ and $\sin\alpha=\frac{p}{2b}$.
Step 6: Use $\cos^2\alpha+\sin^2\alpha=1$.
Then $\frac{p^2}{4a^2}+\frac{p^2}{4b^2}=1$, so $\frac{1}{a^2}+\frac{1}{b^2}=\frac{4}{p^2}$. Comparing with $\frac{1}{a^2}+\frac{1}{b^2}=\frac{k}{p^2}$ gives $k=4$. \[ \boxed{4} \]