= Solution
Use the <electric dipole above a grounded conducting plane> approximation, valid for $d\ll a$. Let $\hat{\mathbf z}$ point upward, and choose $\mathbf p=p(\sin\theta\,\hat{\mathbf x}+\cos\theta\,\hat{\mathbf z})$. Reflecting the two constituent <point charges> and reversing their signs produces an image <electric dipole moment>
$$
\mathbf p'=p(-\sin\theta\,\hat{\mathbf x}+\cos\theta\,\hat{\mathbf z})
$$
at height $-a$. Thus the parallel <electric dipole moment> reverses and the normal one is unchanged. At the real <electric dipole>, the separation from its image is $2a\hat{\mathbf z}$, giving the image <electric field>
$$
\mathbf E_{\rm im}=\frac{1}{4\pi\epsilon_0(2a)^3}
\left[3(\mathbf p'\cdot\hat{\mathbf z})\hat{\mathbf z}-\mathbf p'\right]
=\frac{p}{32\pi\epsilon_0a^3}
\left(\sin\theta\,\hat{\mathbf x}+2\cos\theta\,\hat{\mathbf z}\right).
$$
The physical <torque> is $\boldsymbol\tau=\mathbf p\times\mathbf E_{\rm im}$, so
$$
\boxed{\boldsymbol\tau=-\frac{p^2\sin\theta\cos\theta}{32\pi\epsilon_0a^3}\hat{\mathbf y}
=-\frac{p^2\sin2\theta}{64\pi\epsilon_0a^3}\hat{\mathbf y}.}
$$
For $0<\theta<\pi/2$, its direction reduces $\theta$, tending to align the <electric dipole> with the upward normal. In general it favours either normal orientation; a perfectly parallel orientation has zero <torque> but is unstable to tilting.
The conductor-response <potential energy> includes one half of the real-image interaction:
$$
U=-\frac12\mathbf p\cdot\mathbf E_{\rm im}
=-\frac{p^2(1+\cos^2\theta)}{64\pi\epsilon_0a^3}.
$$
This factor accounts for the induced <surface charge> changing as the physical source is assembled. Differentiation with respect to height at fixed orientation gives
$$
\boxed{\mathbf F=-\frac{\partial U}{\partial a}\hat{\mathbf z}
=-\frac{3p^2(1+\cos^2\theta)}{64\pi\epsilon_0a^4}\hat{\mathbf z}.}
$$
There is no horizontal <force>, since translating the entire arrangement parallel to the plane leaves $U$ unchanged. For example, the normal and parallel orientations give respective downward <force> magnitudes $3p^2/(32\pi\epsilon_0a^4)$ and $3p^2/(64\pi\epsilon_0a^4)$.
A single physical <point charge> induces total plane <electric charge> $-q$. One can verify this by integrating its induced <surface charge density>, $-qh/[2\pi(\varrho^2+h^2)^{3/2}]$, over the plane. The two physical <electric charges> therefore induce opposite totals, giving
$$
\boxed{Q_{\rm plane}=-(+q)-(-q)=0.}
$$
The local <surface charge density> is generally nonzero. The zero total holds for the finite neutral pair as well as its ideal <electric dipole> limit.
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