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 Wheatstones bridge. Hence, the equivalent resistance \[{{R}_{AB}}=\frac{(4+4)\times (4+4)}{(4+4)+(4+4)}\] \[=\frac{8\times 8}{8+8}\] \[=\frac{64}{16}=4\Omega \]