= Solution
The outgoing solution obtained from the <retarded fundamental solution> of the <wave equation> is
$$
h_{ab}(t,\mathbf x)=4\int
\frac{T_{ab}-\tfrac12\eta_{ab}T}
{|\mathbf x-\mathbf x'|}
\left(t-|\mathbf x-\mathbf x'|,\mathbf x'\right)d^3x'.
$$
In the <radiation zone>, put $r=|\mathbf x|$, $\mathbf n=\mathbf x/r$, and retain the leading $1/r$ term at <retarded time> $u=t-r$. The trace term disappears after applying the <transverse-traceless projector>
$$
P_{ij}=\delta_{ij}-n_in_j,
\qquad
\Lambda_{ij,kl}=P_{ik}P_{jl}-\frac12P_{ij}P_{kl},
$$
so
$$
h^{\rm TT}_{ij}(t,\mathbf x)
=\frac4r\Lambda_{ij,kl}(\mathbf n)
\int T_{kl}(u,\mathbf x')d^3x'.
$$
Twice using <stress-energy conservation>, $\partial_aT^{ab}=0$, and integrating by parts gives
$$
\frac{d^2}{du^2}\int T_{00}x_kx_l\,d^3x
=2\int T_{kl}\,d^3x.
$$
The projector removes the trace, so in terms of the <mass quadrupole moment>
$$
Q_{kl}=\int\rho
\left(x_kx_l-\frac13\delta_{kl}|\mathbf x|^2\right)d^3x
$$
the far field is
$$
\boxed{h^{\rm TT}_{ij}=\frac2r\Lambda_{ij,kl}\ddot Q_{kl}(u)}.
$$
The <gravitational-wave energy flux> is
$$
\frac{dP}{d\Omega}
=\frac{r^2}{32\pi}
\left\langle\dot h^{\rm TT}_{ij}\dot h^{\rm TT}_{ij}\right\rangle.
$$
For a trace-free symmetric tensor $A_{ij}$, the <isotropic tensor integral> over the observation direction gives
$$
\int \Lambda_{ij,kl}A_{kl}\Lambda_{ij,mn}A_{mn}\,d\Omega
=\frac{8\pi}{5}A_{ij}A_{ij}.
$$
Consequently the standard <quadrupole formula>, in the units $G=c=1$ used by the paper, is
$$
\boxed{P=\frac15\left\langle\dddot Q_{ij}\dddot Q_{ij}\right\rangle}.
$$
Thus the printed coefficient $4\pi/5$ is inconsistent with the stated definition of $Q_{ij}$ and the standard wave-energy normalization; it appears to be a typographical error.
For the planet, choose its <circular orbit> in the $xy$-plane and write $\omega=2\pi/\tau$. With the star treated as fixed,
$$
\mathbf x=R(\cos\omega t,\sin\omega t,0),
$$
and hence
$$
Q_{ij}=MR^2
\begin{pmatrix}
\cos^2\omega t-\tfrac13&\sin\omega t\cos\omega t&0\\
\sin\omega t\cos\omega t&\sin^2\omega t-\tfrac13&0\\
0&0&-\tfrac13
\end{pmatrix}.
$$
Direct differentiation gives
$$
\dddot Q_{ij}\dddot Q_{ij}=32M^2R^4\omega^6,
$$
so
$$
\boxed{P=\frac{32}{5}M^2R^4\omega^6
=\frac{2048\pi^6}{5}\frac{M^2R^4}{\tau^6}}.
$$
Restoring units multiplies this by $G/c^5$. If the printed $4\pi/5$ coefficient is followed literally, the answer is instead $4\pi$ times larger,
$$
\boxed{P_{\rm printed}=\frac{8192\pi^7}{5}\frac{M^2R^4}{\tau^6}}.
$$
For two bodies of comparable mass, $M$ is replaced by the <reduced mass> and $R$ by their separation.
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