= Solution
Use the sign $s=\pm1$ to label a <point charge> $sq$ at $s\mathbf p/(2q)$, with <velocity> $s\mathbf p'/(2q)$. All quantities on the right below are evaluated at $t_0=t-r/c$. To first order,
$$
\mathbf R_s=\mathbf r-\frac{s\mathbf p}{2q}+O(q^{-2}),\qquad
R_s=r-\frac{s\mathbf n\cdot\mathbf p}{2q}+O(q^{-2}),\qquad
\mathbf v_s=\frac{s\mathbf p'}{2q}+O(q^{-2}).
$$
The denominator in each <Liénard–Wiechert potential> therefore expands as
$$
D_s=R_s-\mathbf R_s\cdot\mathbf v_s/c
=r-\frac{s}{2q}\left(\mathbf n\cdot\mathbf p+\frac{\mathbf r\cdot\mathbf p'}c\right)+O(q^{-2}),
$$
so
$$
\frac1{D_s}=\frac1r+\frac{s}{2qr^2}
\left(\mathbf n\cdot\mathbf p+\frac{\mathbf r\cdot\mathbf p'}c\right)+O(q^{-2}).
$$
In the sum of the two <electric potentials>, the $q/r$ terms cancel and the finite terms add. This gives the <retarded potentials of a point electric dipole>:
$$
\boxed{\phi(\mathbf r,t)=\frac1{4\pi\epsilon_0}
\left(\frac{\mathbf r\cdot\mathbf p(t_0)}{r^3}
+\frac{\mathbf r\cdot\mathbf p'(t_0)}{cr^2}\right),\qquad r>0.}
$$
For the <magnetic vector potential>, the signed <electric charge> and signed <velocity> multiply to give the same sign for both contributions:
$$
\mathbf A=\frac1{4\pi\epsilon_0c^2}\sum_{s=\pm1}\frac{sq\,\mathbf v_s}{D_s}
\longrightarrow\frac{\mathbf p'}{4\pi\epsilon_0c^2r}.
$$
Since $\mu_0\epsilon_0c^2=1$,
$$
\boxed{\mathbf A(\mathbf r,t)=\frac{\mu_0}{4\pi r}\mathbf p'(t_0).}
$$
The $\mathbf p'$ term in the <electric potential> arises from the moving-source denominator and remains finite in the point-<electric dipole> limit; discarding it would fail to give the full retarded <electric field>.
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