\(3a^2+3b^2=10ab\Rightarrow3a^2+3b^2-10ab=0\)

\(\Rightarrow3a^2-9ab-ab+3b^2=0\Rightarrow3a\left(a-3b\right)-b\left(a-3b\right)=0\)

\(\Rightarrow\left(3a-b\right)\left(a-3b\right)=0\)

\(b>a>0\Rightarrow a< 3b\Rightarrow a-3b< 0\).So,3a - b = 0

\(\Rightarrow b=3a\Rightarrow P=\dfrac{a-3a}{a+3a}=\dfrac{-2a}{4a}=-\dfrac{1}{2}\)