For a Wheatstone bridge to be balanced, the ratio of resistances in adjacent arms must be equal. Based on the values, the condition is $\frac{P}{Q} = \frac{S_{eq}}{R}$ (or equivalent $\frac{P}{S_{eq}} = \frac{Q}{R}$). Substituting the known values: $\frac{2}{3} = \frac{S_{eq}}{6} \implies S_{eq} = \frac{12}{3} = 4 \, \Omega$. The existing resistance is $S = 8 \, \Omega$. To reduce this to $4 \, \Omega$, a shunt resistance $R_{sh}$ must be connected in parallel. Using the parallel resistance formula: $\frac{1}{S_{eq}} = \frac{1}{S} + \frac{1}{R_{sh}} \implies \frac{1}{4} = \frac{1}{8} + \frac{1}{R_{sh}} \implies \frac{1}{R_{sh}} = \frac{1}{8}$. Thus, $R_{sh} = 8 \, \Omega$ .