 Solution

Joining the two middle nodes makes the upper lower-network transistors, controlled by $a,c$, parallel with each other. The lower transistors, controlled by $b,d$, are also parallel. These two parallel groups are in series, so the lower conducting condition becomes $(a\vee c)\wedge (b\vee d)$.

Its dual upper network must be two parallel branches, one containing the series PMOS transistor pair $a,c$, the other the series pair $b,d$. This is the right panel of the original drawing. Its condition is $(\neg a\wedge \neg c)\vee (\neg b\wedge \neg d)$. Merely leaving the old upper network unchanged would break CMOS complementarity for some inputs, allowing a supply short rather than defining the required logic.