$\text{ The smallest path difference for point P to be a maxima is }\lambda$

$\Rightarrow \sqrt D^2 + x^2 - \sqrt {(D-3\lambda)^2 + x^2} = \lambda$

$ (D-3\lambda)^2 + x^2 = \lambda^2 + D^2 + x^2 - 2\lambda (\sqrt D^2 + x^2)$

$\Rightarrow 9\lambda^2 -6\lambda D = \lambda^2 - 2\lambda (\sqrt D^2 + x^2)$

$ \text{we can ignore }\lambda^2 \text{ as compared to }\lambda D$

$\Rightarrow  3 D =  \lambda (\sqrt D^2 + x^2)$

$\text{Squaring both sides we get} x^2 = 8D^2 , \Rightarrow x = 2\sqrt 2 D$