Cho số phức $z=a+bi,\left( a;b\in \mathbb{R} \right)$ thỏa mãn...

Cách 1: (Môđun hóa hai vế)
$\begin{aligned}
& \overline{z}+1+8i-\left( 1+i \right)\left| z \right|=0\Leftrightarrow \overline{z}=\left| z \right|-1+\left( \left| z \right|-8 \right)i\Rightarrow {{\left| \overline{z} \right|}^{2}}={{\left( \left| z \right|-1 \right)}^{2}}+{{\left( \left| z \right|-8 \right)}^{2}} \\
& \Leftrightarrow {{\left| z \right|}^{2}}=2{{\left| z \right|}^{2}}-18\left| z \right|+65\Leftrightarrow {{\left| z \right|}^{2}}-18\left| z \right|+65=0\Leftrightarrow \left[ \begin{aligned}
& \left| z \right|=5 \\
& \left| z \right|=13 \\
\end{aligned} \right. \\
\end{aligned}$
Mà $\left| z \right|>6$. Vậy $\left| z \right|=13\Rightarrow \overline{z}=12+5i\Rightarrow z=12-5i\Rightarrow a=12,b=-5\Rightarrow a+2b=2.$
Cách 2: Ta có: $\overline{z}+1+8i-\left( 1+i \right)\left| z \right|=0\Leftrightarrow a-bi+1+8i-\left( 1+i \right)\left( \sqrt{{{a}^{2}}+{{b}^{2}}} \right)=0$
$\begin{aligned}
& \Leftrightarrow \left( a+1-\sqrt{{{a}^{2}}+{{b}^{2}}} \right)+\left( 8-b-\sqrt{{{a}^{2}}+{{b}^{2}}} \right)i=0 \\
& \Leftrightarrow \left\{ \begin{aligned}
& a+1-\sqrt{{{a}^{2}}+{{b}^{2}}}=0 \\
& 8-b-\sqrt{{{a}^{2}}+{{b}^{2}}}=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a+b-7=0 \\
& a+1-\sqrt{{{a}^{2}}+{{b}^{2}}}=0 \\
\end{aligned} \right. \\
\end{aligned}$
$\Leftrightarrow \left\{ \begin{aligned}
& b=7-a \\
& a+1-\sqrt{{{a}^{2}}+{{\left( 7-a \right)}^{2}}}=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& b=7-a \\
& \left[ \begin{aligned}
& a=12 \\
& a=4 \\
\end{aligned} \right. \\
\end{aligned} \right.$
Vậy $a=4\Rightarrow b=3$ khi đó $\left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}=5$ không thỏa mãn điều kiện
Với $a=12\Rightarrow b=-5$ khi đó $\left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}=13$ thỏa mãn điều kiện.
Vậy $P=a+2b=2$