= 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 $(a\lor c)\land(b\lor d)$.
Its dual upper network must be \b[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 $(\neg a\land\neg c)\lor(\neg b\land\neg d)$. Merely leaving the old upper network unchanged would break <CMOS> complementarity for some inputs, allowing a supply short rather than defining the required logic.
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