Perpendicular line from $P$ to $Q$ intersects $AD$ at $Q$. We have $QA=QP=\dfrac{AP}{\sqrt2}$, and $$\dfrac{FA}{AE}=\dfrac{FQ}{QP}=\dfrac{FA-QA}{QP}\Rightarrow\dfrac{QP}{AE}=\dfrac{FA-QA}{FA}=\dfrac{FA-QP}{FA}=1-\dfrac{QP}{FA}\Rightarrow\dfrac{QP}{AE}+\dfrac{QP}{FA}=1$$$$\dfrac{AP}{AE}+\dfrac{AP}{FA}=\sqrt2\Rightarrow\dfrac{1}{AE} + \dfrac{1}{AF} = \dfrac{\sqrt2}{AP}$$ b) is obtained by $GM-HM$ inequality. Simply, $$\sqrt{AE\cdot AF}\geq \dfrac{2}{\dfrac{1}{AE}+\dfrac{1}{AF}}= \dfrac{2}{\dfrac{\sqrt2}{AP}} = AP\sqrt2\Rightarrow AE\cdot AF\geq 2AP^2\Rightarrow \dfrac{AE\cdot AF}{2}\geq AP^2 $$