The circuit can be shown as given below
The equivalent resistance between $D$ and $C$
$R_{D C} =\frac{15 \times(15+15)}{15+(15+15)}=\frac{15 \times 30}{15+30} $
$=\frac{15 \times 30}{45}=10\, \Omega$
Now, between $A$ and $B$, the resistance of upper part $A D C B$,
$R_{1}=15+10+15=40 \,\Omega$
Between $A$ and $B$, the resistance of middle part $A O B$
$R_{2}=15+15=30\, \Omega$
Therefore, equivalent resistance between $A$ and $B$
$\frac{1}{R'} =\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}=\frac{1}{40}+\frac{1}{30}+\frac{1}{15} $
$=\frac{3+4+8}{120}=\frac{15}{120} $
$\Rightarrow R'=\frac{120}{15}=8\, \Omega$
