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The two lower conducting branches are the series pairs controlled by $a,b$ and by $c,d$. Their overall NMOS transistor conduction condition is $ab\vee cd$. Apply complementary CMOS network duality: the upper network has the parallel pair of PMOS transistors controlled by $a,b$, in series with the parallel pair controlled by $c,d$. Its first end connects to the high rail, its last end connects to $z$, and the junction between the two pairs is internal to the upper network.

The left panel gives that completion. Every upper transistor uses its correspondingly named input and conducts when that input is low. Its conducting condition is $(\neg a\vee \neg b)\wedge (\neg c\vee \neg d)$, exactly the complement of the lower condition. Thus the ideal static circuit connects the output to one supply rail for every stable binary input.

Factor the common input before constructing the networks. Use an NMOS transistor controlled by $b$ in series with parallel NMOS transistors controlled by $a$ and $c$. Its conducting condition remains $b(a\vee c)$. The complementary upper network is a PMOS transistor controlled by $b$ in parallel with a series chain of PMOS transistors controlled by $a$ and $c$; its conducting condition is $\neg b\vee \neg a\neg c$.

This is the right panel in part (a). It uses three lower plus three upper transistors, six in total instead of eight, and needs no separate inverter or complemented external input. The complementary CMOS network duality ensures the same static Boolean function and complementary rail connections.