$(1+i)z+2\overline{z}=3+2i\\ <=>(1+i)(a+bi)+2(a-bi)=3+2i\\ <=>a+bi+ai-b+2a-2bi=3+2i\\ <=>3a-b+(a-b)i=3+2i\\ =>\left\{\begin{array}{l} 3a-b=3\\ a-b=2\end{array} \right.\\ <=>\left\{\begin{array}{l} a=\dfrac{1}{2}\\ b=\dfrac{-3}{2}\end{array} \right.\\ =>P=a+b=\dfrac{1}{2}+\dfrac{-3}{2}=-1$