We have the following cases :

$$\bullet$$ For configuration (I) : Considering as Wheatstone's bridge, we get the equivalent resistance as $${({R_I})_{eq}} = 1\,\Omega $$.

$$\bullet$$ For configuration (II) : The given configuration is further reduced as follows:

That is, $${({R_{II}})_{eq}} = {1 \over 2}\,\Omega $$.

$$\bullet$$ For configuration (III) : Similarly, we get $${({R_{III}})_{eq}} = 2\,\Omega $$.

The power is given by

$$P = {{{V^2}} \over R}$$

Since V is the same for all three configurations, we have the power as a quantity indirectly proportional to the resistance :

$$P \propto {1 \over R}$$

From the above three case, we have $${({R_I})_{eq}} = 1\,\Omega ;\,{({R_{II}})_{eq}} = {1 \over 2}\,\Omega ;\,{({R_{III}})_{eq}} = 2\,\Omega $$, which implies that

$${R_{II}} > {R_I} > {R_{III}}$$

Therefore, $${P_2} > {P_2} > {P_3}$$