\(\sqrt{ab}\le\frac{a+b}{2}\Rightarrow\frac{1}{\sqrt{ab}}\ge\frac{2}{a+b}\)
\(\Rightarrow P\ge\frac{2}{a+b}+\frac{1}{b}=\frac{2}{a+b}+\frac{2}{2b}=2\left(\frac{1}{a+b}+\frac{1}{2b}\right)\ge2.\frac{4}{a+3b}\ge8\)
\(\Rightarrow P_{min}=8\) khi \(a=b=\frac{1}{4}\)