= Solution
The modified output is
$$
\boxed{z'=\neg((a\lor c)\land(b\lor d))
=(\neg a\land\neg c)\lor(\neg b\land\neg d).}
$$
For example, \b[$(a,b,c,d)=(1,0,0,1)$ gives $z=1$ and $z'=0$]. Neither original series branch is complete, but the new connection supplies the cross-route through the transistors controlled by $a,d$. Interchanging those roles gives the other cross-route example. In fact $(a\lor c)(b\lor d)=ab\lor ad\lor bc\lor cd$, so the modified circuit has added exactly those two possible lower conduction routes.
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