Here's how you do it.

First, notice that for any state $|\psi\rangle$, we have
\begin{align}
  \langle p|[\hat x, \hat p]|\psi\rangle
  &= \langle p|\hat x\hat p-\hat p\hat x|\psi\rangle \\
  &= \langle p|\hat x\hat p|\psi\rangle - \langle p|\hat p \hat x|\psi\rangle \\
  &= \langle p|\hat x \hat p|\psi\rangle - p\langle p|\hat x|\psi\rangle
\end{align}
Now, recall the canonical commutation relation between $\hat x$ and $\hat p$;
\begin{align}
  [\hat x, \hat p] &= i\hbar \hat I
\end{align}
where $\hat I$ is the identity operator.  Using this fact on the left hand side of the manipulation we just performed, and doing a little rearranging, we find that
\begin{align}
  p\langle p|\hat x|\psi\rangle = \langle p|\hat x \hat p|\psi\rangle -i\hbar\langle p|\psi\rangle. \tag{1}
\end{align}
Now, focus on the first term on the right.  we have
\begin{align}
  \langle p|\hat x \hat p|\psi\rangle
  &= \int dx\,\langle p|\hat x|x\rangle\langle x|\hat p|\psi\rangle \\
  &= \int dx\, x e^{ipx/\hbar}\left[-i\hbar \frac{d}{dx}\psi(x)\right] \\
  &= i\hbar \int dx\, \frac{d}{dx}(xe^{ipx/\hbar})\psi(x) \\
  &= i\hbar \int dx\, e^{ipx/\hbar}\psi(x) + i\hbar\int dx\,x\left(\frac{ip}{\hbar}\right)e^{ipx/\hbar}\psi(x) \\
  &= i\hbar\psi(p) -p\int dx\,xe^{ipx/\hbar}\psi(x)\\
  &= i\hbar\psi(p)+i\hbar p\frac{d}{dp}\int dx\,e^{ipx/\hbar}\psi(x) \\
  &= i\hbar\psi(p)+i\hbar p \frac{d}{dp}\psi(p) \tag{2}
\end{align}
where we have made liberal use of integration by parts, and the following identities:
\begin{align}
  \int dx\, |x\rangle\langle x| = \hat I, \qquad \psi(x) \overset{\mathrm{def}}{=} \langle x|\psi\rangle, \qquad \psi(p) \overset{\mathrm{def}}{=} \langle p|\psi\rangle,\qquad \langle x|\hat p|\psi\rangle = -i\hbar\frac{d}{dx}\psi(x).
\end{align}
The last fact is precisely the statement that the position space representation of the momentum operator is $-i\hbar d/dx$. Now combine $(1)$ and $(2)$ to obtain
\begin{align}
  p\langle p|\hat x|\psi\rangle = i\hbar p\frac{d}{dp}\psi(p) \tag{3}
\end{align}
Next, we simply note that if we denote the momentum space representation of the position operator as $D^{(p)}(\hat x)$, then by definition
\begin{align}
  D^{(p)}(\hat x)\psi(p) = \langle p|\hat x|\psi\rangle,
\end{align}
Combining this with $(3)$, and noting that the resulting equation should hold for all $p$, and dividing both sides by $p$, we obtain the desired result;
\begin{align}
  D^{(p)}(\hat x) = i\hbar \frac{d}{dp}
\end{align}