The equivalent resistance across A and B is

The equivalent circuit can be redrawn as We have, $\frac{P}{Q}=\frac{R}{S}$ ie, $\frac{4}{4}=\frac{4}{4}$ So, the given circuit is a balanced Wheatstone's bridge. Hence, the equivalent resistance $R_{A B}=\frac{(4+4) \times(4+4)}{(4+4)+(4+4)}=\frac{8 \times 8}{8+8}=\frac{64}{16}=4 \Omega$