Seven resistances are connected as shown in the figure. The equivalent resistance between A and B is

The given circuit can be simplified as follows:

Now it is a balanced Wheatstone bridge.
So,

$\Rightarrow {\mathrm{R}}_{\mathrm{AB}}=\frac{8\times 8}{8+8}=\frac{64}{16}=4\mathrm{\Omega }$