= Solution
The leading <electric dipole radiation> fields are
$$
\mathbf E_{\rm rad}=\frac{\mathbf n(\mathbf n\cdot\mathbf p'')-\mathbf p''}{4\pi\epsilon_0c^2r},\qquad
\mathbf B_{\rm rad}=\frac{\mu_0}{4\pi cr}\mathbf p''\times\mathbf n
=\frac1c\mathbf n\times\mathbf E_{\rm rad}.
$$
The radiation <electric field> is transverse, $\mathbf n\cdot\mathbf E_{\rm rad}=0$. The <vector triple product> therefore gives the leading <Poynting vector>
$$
\boldsymbol{\mathcal P}=\frac1{\mu_0c}\mathbf E_{\rm rad}\times(\mathbf n\times\mathbf E_{\rm rad})+O(r^{-3})
=\frac{|\mathbf E_{\rm rad}|^2}{\mu_0c}\mathbf n+O(r^{-3}).
$$
If $\theta$ is the angle between $\mathbf p''(t-r/c)$ and $\mathbf n$, then
$$
|\mathbf n(\mathbf n\cdot\mathbf p'')-\mathbf p''|^2
=|\mathbf p''|^2-(\mathbf n\cdot\mathbf p'')^2
=|\mathbf p''|^2\sin^2\theta.
$$
Using $\mu_0\epsilon_0c^2=1$, the final result is
$$
\boxed{\boldsymbol{\mathcal P}(\mathbf r,t)=
\frac{|\mathbf p''(t-r/c)|^2\sin^2\theta}{16\pi^2\epsilon_0c^3r^2}\mathbf n+O(r^{-3}),\qquad
|\boldsymbol{\mathcal P}|=\frac{|\mathbf p''(t-r/c)|^2\sin^2\theta}{16\pi^2\epsilon_0c^3r^2}+O(r^{-3}).}
$$
The leading <Poynting flux> is outward, vanishes along the instantaneous retarded $\mathbf p''$ axis, and is maximal in its perpendicular plane. If $\mathbf p''=0$, the radiation term is zero and no angle needs to be assigned. The large-$r$ expansion assumes bounded retarded <electric dipole moment> and time <derivatives>; it may also be read at fixed <retarded time>.
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