\(2\sqrt{ab}\le a+b\le4\Rightarrow\sqrt{ab}\le2\Rightarrow ab\le4\Rightarrow\frac{1}{ab}\ge\frac{1}{4}\)
\(P=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{16}{ab}+ab+\frac{17}{2ab}\)
\(P\ge\frac{4}{\left(a+b\right)^2}+2\sqrt{\frac{16ab}{ab}}+\frac{17}{2}.\frac{1}{4}\ge\frac{4}{4^2}+\frac{81}{8}=\frac{83}{8}\)
\(\Rightarrow P_{min}=\frac{83}{8}\) khi \(a=b=2\)