Determine the rate equation (intermediate reaction)
You can't write $[\ce{B}] = [\ce{C}]$ because  $\ce{C}$ is consumed in the second step but $\ce{B}$ isn't. However, you were pretty much on the right track.

Since $\ce{E}$ is produced from $\ce{D}$, you can write $[\ce{E}] = [\ce{D}]_i - [\ce{D}]$ where $[\ce{X}]$ denotes the concentration of $\ce{X}$ at a certain point in time and $[\ce{X}]_i$ denotes the initial concentration of $\ce{X}$. This is assuming that the initial concentration of $\ce{E}$ is $0$, of course.

On top of this, $\ce{C}$ is either produced from $\ce{A}$, or consumed in the production of $\ce{E}$. Therefore you can write $[\ce{C}] = [\ce{A}]_i - [\ce{A}] - [\ce{E}] = [\ce{A}]_i - [\ce{A}] - [\ce{D}]_i + [\ce{D}]$. (There is the reverse step in which $\ce{C}$ produces $\ce{A}$, but the equation still holds, since you can just consider this to be "negative" production of $\ce{C}$ from $\ce{A}$.)

Your rate law is therefore: $r = k_1[\ce{D}]\big([\ce{A}]_i - [\ce{A}] - [\ce{D}]_i + [\ce{D}]\big)$. The initial concentrations are merely constants, so you could view this as being of the form $r = k_a[\ce{D}] + k_b[\ce{A}][\ce{D}] + k_c[\ce{D}]^2$. I can't think of any way to simplify this (the steady-state approximation is not very valid here since $\ce{C}$ is produced quickly and consumed slowly, and in any case it does not yield a nicer expression). I imagine that one could use some form of an integrated rate law to find expressions for $[\ce{A}]$ and/or $[\ce{D}]$, but it would just make things even uglier.