The quantity $\lambda(P)|P|$ is scale invariant since $\lambda(tP)=\frac{1}{t^2}\lambda(P)$. Therefore, its derivative at $t=1$ equals zero.
The derivative of the eigenvalue can be found using the shape derivative formula:
$$\lambda'(P)(\theta) = -\int_{\partial P} (\partial_n u)^2 (\theta \cdot n)$$
It is immediate to notice that a vector field which corresponds to dilations verifies $\theta \cdot n = 1$.
In the same case the shape derivative of the area gives
$$ |P|'(\theta) = \int_{\partial P} \theta \cdot n.$$
This gives $\lambda(P)|P|'=-|P|\lambda'(P)(\theta) = |P|\int_{\partial P}(\partial_n u)^2$.
In the end: $\lambda(P)=|P| \frac{1}{|\partial P|}\int_{\partial P}(\partial_n u)^2$.
If my reasoning is correct, this is an equality (always better than an inequality :) )
Therefore the inequality the OP wants goes in the other sense (the average is smaller than the $L^\infty$ norm).