Determine the rate law for the overall reaction (where the overall rate constant is represented as k)?

I would get that #r(t) ~~ k[A][D]#.

#A stackrel(k_1" ")rightleftharpoons B + C# #" "" "#(#"fast equilibrium"#)
#""^(" "k_(-1))#
#ul(C + D stackrel(k_2" ")(->) E" "" ")# (#"slow"#)
#A + D stackrel(k" ")(->) B + E#

We begin with the notion that the slow step is the rate-determining step. From that, the initial form of the rate law is:

#r(t) = k_2[C][D]#

where #k_2# is the rate constant for the forward second step.

However, #C# is not a reactant (but is an intermediate), and as such, this is not an appropriate rate law yet. Instead, we should recognize that

#K = k_1/k_(-1) = ([B][C])/([A])#

is the equilibrium constant for the first reaction step. Therefore:

#[C] = k_1/k_(-1) ([A])/([B])#

and we can substitute back into the rate law to get:

#r(t) ~~ (k_2k_1)/(k_(-1)[B]) [A][D]#

We would use #[B]# at equilibrium, so the approximation is that #[B]# is a constant after step 1 and stays constant through step 2 in the reaction mechanism (because it isn't a reactant or product in step 2).

In the expected notation,

#color(blue)(r(t) ~~ k[A][D])#,

where #k = (k_2k_1)/(k_(-1)[B])#.