Show that the electric field of a (perfect) dipole (see Eq. 3.103) can be written in the coordinate-free form:
\[\mathbf{E}_{\text{dip} }(\mathbf{r})=\frac{1}{4\pi \epsilon _0 }\frac{1}{r^3 } \left[ 3(\mathbf{p}\cdot \hat{\mathbf{r}})\hat{\mathbf{r}}-\mathbf{p} \right] \tag{Eq. 3.104}\]Let's start by recalling Eq. 3.103 as found on section 3.4.4: The Electric Field of a Dipole:
\[\mathbf{E}_{\text{dip} }(r,\theta )=\frac{p}{4\pi \epsilon _0 r^3}\bigg( 2\cos \theta \hat{\mathbf{r}} + \sin \theta \hat{\boldsymbol{\theta }} \bigg) \tag{Eq. 3.103, page 155}\]There are a few ways to jump from 3.103 to 3.104 and vice versa. My favorite is through the definition of dot products as projections. Any vector, say $\mathbf{a}=(a_x,a_y) $, can be written as:
\[\mathbf{a}=a_x\hat{\mathbf{x}}+a_y\hat{\mathbf{y}} \]So far this is just notation. Now, what is $a_x $? It is the $x $-component of $\mathbf{a} $. We find this component through a projection:
\[a_x=\mathbf{a}\cdot \hat{\mathbf{x}} \]In this sense, we write:
\[\mathbf{p}=p_r\hat{\mathbf{r}} +p_{\theta } \hat{\boldsymbol{\theta} }=(\mathbf{p}\cdot \hat{\mathbf{r}})\hat{\mathbf{r}}+(\mathbf{p}\cdot \hat{\boldsymbol{\theta} })\hat{\boldsymbol{\theta} } \]Replace $\mathbf{p} $ in Eq. 3.104:
\begin{align*} \mathbf{E}_{\text{dip} }(\mathbf{r}) &= \frac{1}{4\pi \epsilon _0 }\frac{1}{r^3 } \left[ 3(\mathbf{p}\cdot \hat{\mathbf{r}})\hat{\mathbf{r}}-(\mathbf{p}\cdot \hat{\mathbf{r}})\hat{\mathbf{r}}-(\mathbf{p}\cdot \hat{\boldsymbol{\theta} })\hat{\boldsymbol{\theta} } \right] \\ &= \frac{1}{4\pi \epsilon _0 }\frac{1}{r^3 } \left[ 2(\mathbf{p}\cdot \hat{\mathbf{r}})\hat{\mathbf{r}}-(\mathbf{p}\cdot \hat{\boldsymbol{\theta} })\hat{\boldsymbol{\theta} } \right] \end{align*}We now use the fact
\begin{align*} \mathbf{p}\cdot \hat{\mathbf{r}} &= p \cos \left( \theta \right) \\ \mathbf{p}\cdot \hat{\boldsymbol{\theta }} &= -p \sin \left( \theta \right) \end{align*}And substitute above:
\begin{align*} \mathbf{E}_{\text{dip} }(\mathbf{r}) &= \frac{1}{4\pi \epsilon _0 }\frac{1}{r^3 } \left[ 2(\mathbf{p}\cdot \hat{\mathbf{r}})\hat{\mathbf{r}}-(\mathbf{p}\cdot \hat{\boldsymbol{\theta} })\hat{\boldsymbol{\theta} } \right] \\ &= \frac{1}{4\pi \epsilon _0 }\frac{1}{r^3 } \left[ 2(p\cos \left( \theta \right) )\hat{\mathbf{r}}-(-p \sin \left( \theta \right) )\hat{\boldsymbol{\theta} } \right] \\ &= \frac{1}{4\pi \epsilon _0 }\frac{1}{r^3 } \left[ 2(p\cos \left( \theta \right) )\hat{\mathbf{r}}+(p \sin \left( \theta \right) )\hat{\boldsymbol{\theta} } \right]\\ &= \frac{p}{4\pi \epsilon _0 }\frac{1}{r^3 } \left[ 2(\cos \left( \theta \right) )\hat{\mathbf{r}}+( \sin \left( \theta \right) )\hat{\boldsymbol{\theta} } \right] \end{align*}Which is precisely Eq. 3.103. If we want to go from $3.103 $ to $3.104 $, we can use the same relations for $\mathbf{p}\cdot \hat{\mathbf{r}} $ and $\mathbf{p}\cdot \hat{\boldsymbol{\theta }} $ and build up from there.
If you are confused as to how to get the equations for $\mathbf{p}\cdot \hat{\mathbf{r}} $ and $\mathbf{p}\cdot \hat{\boldsymbol{\theta }} $, consider the following:
The magnitude of $\hat{\mathbf{r}} $ and $\hat{\boldsymbol{\theta }} $ is 1, of course. Further, notice how $\hat{\boldsymbol{\theta }} $ tends to point opposite to $\mathbf{p} $, that's why the ($-$) sign (the dashed green vector is just $\hat{\boldsymbol{\theta }} $ starting at the origin rather at its familiar location (at the tip of $\hat{\mathbf{r}} $)).