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

$\dfrac{\left( \left| z \right|-1 \right)\left( 1+iz \right)}{z-\dfrac{1}{\overline{z}}}=i\Leftrightarrow \dfrac{\left( \left| z \right|-1 \right)\left( 1+iz \right)\overline{z}}{z\overline{z}-1}=i\text{ }\left( \left| z \right|\ne 1 \right)$
$\Leftrightarrow \dfrac{\left( \left| z \right|-1 \right)\left( 1+iz \right)\overline{z}}{{{\left| z \right|}^{2}}-1}=i\Leftrightarrow \dfrac{\left( 1+iz \right)\overline{z}}{\left| z \right|+1}=i$
$\Leftrightarrow \overline{z}+i{{\left| z \right|}^{2}}=i\left( \left| z \right|+1 \right)\Leftrightarrow a-bi+\left( {{a}^{2}}+{{b}^{2}} \right)i=i\left( \sqrt{{{a}^{2}}+{{b}^{2}}}+1 \right)$
$\Leftrightarrow a+\left( -b+{{a}^{2}}+{{b}^{2}} \right)i=i\left( \sqrt{{{a}^{2}}+{{b}^{2}}}+1 \right)\Leftrightarrow \left\{ \begin{aligned}
& a=0 \\
& {{b}^{2}}-b=\left| b \right|+1 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& a=0 \\
& \left[ \begin{aligned}
& \left\{ \begin{aligned}
& b<0 \\
& b=\pm 1\left( loại \right) \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& b>0 \\
& {{b}^{2}}-2b-1=0 \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& a=0 \\
& \left[ \begin{aligned}
& b=1+\sqrt{2}\left( nhận \right) \\
& b=1-\sqrt{2}\left( loại \right) \\
\end{aligned} \right. \\
\end{aligned} \right.$.
Vậy $P=a+b=1+\sqrt{2}.$