Ignoring the normalization, your inequality writes $$\lambda(P)\|u\|_{L^2(P)}^2\ge\|\partial_\nu u\|_{L^\infty(P)}^2|P|\qquad?$$$$\lambda(P)\|u\|_{L^2(P)}^2\ge\|\partial_\nu u\|_{L^\infty(\partial P)}^2|P|\qquad?$$ 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.Perhaps also you should think to an inequality of the form $$\lambda(P)\|u\|_{L^2(P)}\ge c\|\partial_\nu u\|_{L^\infty(\partial P)},$$ which is scaling invariant too. For instance the limit case of the unit disk gives $u(x,y)=a(r)$ with $$a''+\frac1r\,a'=-\lambda a \quad a'(0)=0,\quad a(1)=0.$$ Then \begin{eqnarray*} \|\partial_\nu u\|_{L^\infty(\partial P)} & = & |a'(1)|=\left|\int_0^1(ra'(r)'dr\right| \\ & = & \lambda\left|\int_0^1ra(r)dr\right|\le\frac\lambda{2\sqrt\pi}\,\|u\|_{L^2(P)}. \end{eqnarray*}