= Solution
The <electric potential> of a <point charge> $Q$ at the origin is
$$
\boxed{\Phi(\mathbf x)=\frac{Q}{4\pi\epsilon_0|\mathbf x|}}.
$$
For the two charges forming the dipole,
$$
\Phi(\mathbf x)=\frac{Q}{4\pi\epsilon_0}
\left(\frac1{|\mathbf x|}-\frac1{|\mathbf x+\mathbf d|}\right).
$$
The first-order <Taylor expansion> at large $|\mathbf x|$ is
$$
\frac1{|\mathbf x+\mathbf d|}
=\frac1r-\frac{\mathbf d\cdot\mathbf x}{r^3}
+O\!\left(\frac{|\mathbf d|^2}{r^3}\right),
$$
and therefore, with the <electric dipole moment> $\mathbf p=Q\mathbf d$,
$$
\boxed{\Phi(\mathbf x)=\frac1{4\pi\epsilon_0}
\frac{\mathbf p\cdot\mathbf x}{r^3}}
$$
to leading order. Taking the interaction of a second dipole with the corresponding <electric field> gives the stated <electric dipole-dipole interaction>
$$
U=\frac1{8\pi\epsilon_0}
\left(\frac{\mathbf p_1\cdot\mathbf p_2}{r^3}
-\frac{3(\mathbf p_1\cdot\mathbf r)(\mathbf p_2\cdot\mathbf r)}{r^5}\right).
$$
Let the lattice spacing be $d$ and write the central dipole as
$$
\mathbf p=p(\cos\theta,\sin\theta).
$$
Its two horizontal neighbours have moment $p(\cos\theta,-\sin\theta)$. For either one, the expression in parentheses, after extracting $d^{-3}$, is
$$
-p^2(1+\cos^2\theta).
$$
Its two vertical neighbours have moment $p(-\cos\theta,\sin\theta)$, and each contributes instead
$$
-p^2(1+\sin^2\theta).
$$
Adding all four nearest-neighbour interactions and using the <Pythagorean trigonometric identity> gives
$$
\boxed{U_{\mathrm{nearest}}
=-\frac{6p^2}{8\pi\epsilon_0d^3}
=-\frac{3p^2}{4\pi\epsilon_0d^3}}.
$$
The angle has cancelled, so the energy is independent of $\theta$.
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