已知$P(A)=\dfrac{1}{5}$，$P(B|A)=\dfrac{1}{2}$，$P(\overline{B}|\overline{A})=\dfrac{5}{8}$，则$P\left(B\right)= $ $(\qquad)$ $ $.

$\\dfrac{2}{5}$

$\\dfrac{1}{6}$

$\\dfrac{1}{5}$

$\\dfrac{3}{8}$

$\because P(A)=\dfrac{1}{5}$，$P(B|A)=\dfrac{1}{2}$， 
$\therefore P\left(AB\right)=P\left(A\right)P\left(B|A\right)=\dfrac{1}{5}×\dfrac{1}{2}=\dfrac{1}{10}$， 
$\because P(\overline{A})=1-P\left(A\right)=\dfrac{4}{5}$，$P(\overline{B}|\overline{A})=\dfrac{5}{8}$， 
$\therefore P(\overline{A}\overline{B})=P(\overline{A})P(\overline{B}|\overline{A})=\dfrac{4}{5}×\dfrac{5}{8}=\dfrac{1}{2}$， 
$\therefore P\left(A\cup B\right)=1-P(\overline{A}\overline{B})=\dfrac{1}{2}$， 
又$\because P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(AB\right)$， 
$\therefore P\left(B\right)=P\left(A\cup B\right)+P\left(AB\right)-P\left(A\right)=\dfrac{1}{2}+\dfrac{1}{10}-\dfrac{1}{5}=\dfrac{2}{5}$. 
故选：$\rm A $