Given a pull-up network, draw the complimentary pull-down network (CMOS gates)
The problem has been given to you in a deliberately confusing way. Redraw the leftmost four transistors as A in parallel with B and D in parallel with  E then join the parallel pairs with a single line. A general formal approach that will avoid confusion is to write out the boolean expression for your high side circuit, The inputs are inverted because they feed active low input p channel fets.:-

\$Y=\bar{C}.\bar{F}+(\bar{A}+\bar{B}).(\bar{D}+\bar{E})\$

Now you need the inverse of this for Y low:-

\$ \bar{Y}=\overline{\bar{C}.\bar{F}+(\bar{A}+\bar{B}).(\bar{D}+\bar{E})}\$

You also need to get to the un-inverted inputs for the active high input n channel fets using the transforms:-

\$\overline{S+T}=\bar{S}.\bar{T}\$

\$\overline{S.T}=\bar{S}+\bar{T}\$

This is the formal version of switch parallel and series so.

\$ \bar{Y}=\overline{\bar{C}.\bar{F}+(\bar{A}+\bar{B}).(\bar{D}+\bar{E})}\$

\$ \bar{Y}=\overline{\overline{C+F}+\overline{A.B}.\overline{D.E}}\$

\$ \bar{Y}=\overline{\overline{C+F}+\overline{A.B+D.E}}\$

\$ \bar{Y}=(C+F).(A.B+D.E)\$

Draw this out as transistors and you have your answer.