 Solution

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.