Ignoring the normalization, your inequality writes $$\lambda(P)\|u\|_{L^2(P)}^2\ge\|\partial_\nu u\|_{L^\infty(P)}^2|P|\qquad?$$ Good news: this is scaling invariant.
Bad news: this is false for a square $P=(-\pi/2,\pi/2)^2$. Then $u(x,y)=\cos x\cos y$,$\lambda(P)=2$, $|P|=\pi^2$ and $$\|u\|_{L^2}^2=\left(\int_{-\pi/2}^{\pi/2}\cos^2 x\,dx\right)^2=\frac{\pi^2}4\,,\qquad\|\partial_\nu u\|_{L^\infty}^2=1.$$
Perhaps your inequality is correct with an extra constant factor.