= Solution
Write $P=P_0+dP_1+O(d^2)$ and $r=r_0+dr_1+O(d^2)$, with each coefficient satisfying the zero <Dirichlet boundary condition>. The steady complex equation, including the fluctuating coupling, is $0=\mathcal L_rP-dB\sin(2\pi x/\ell)$. Since $B_0=(P_0-P_0^*)/(2i)$, its order-$d$ terms give
$$
\mathcal L_{r_0}P_1=-ir_1P_0'+\frac{P_0-P_0^*}{2i}\sin(2\pi x/\ell).
$$
This is a <regular perturbation> equation for the threshold shift. The right projection uses the complex <inner product>, not a bilinear product without conjugation.
The <linear operator> $\mathcal L_{r_0}$ is self-adjoint on the domain $H^2(0,\ell)\cap H_0^1(0,\ell)$ in complex <L2 space>. For functions $v,w$ in that domain, two <integration by parts> steps give
$$
\int_0^\ell v^*\mathcal L_{r_0}w\,dx
=\int_0^\ell(\mathcal L_{r_0}v)^*w\,dx
+\left[v^*w'-(v^*)'w+ir_0v^*w\right]_0^\ell.
$$
The boundary term vanishes. In particular $ir_0\partial_x$ is formally self-adjoint: both complex conjugation of $i$ and integration by parts contribute a minus sign. Taking $v=P_0$, $w=P_1$ and using $\mathcal L_{r_0}P_0=0$ proves
$$
\int_0^\ell P_0^*\mathcal L_{r_0}P_1\,dx=0.
$$
This is the <Fredholm solvability condition for a self-adjoint operator>. Define $s(x)=\sin(2\pi x/\ell)$ and $N=\int_0^\ell|P_0|^2\,dx$. The gauged sine mode has
$$
I=\int_0^\ell P_0^*P_0'\,dx=-\frac{ir_0}{2}N,
$$
since the real envelope contributes an integral of its own derivative, whose endpoint term is zero. In particular the denominator in the solvability condition is nonzero. Therefore
$$
\boxed{r_1=-\frac{\displaystyle\int_0^\ell P_0^*(P_0-P_0^*)s(x)\,dx}{\displaystyle2\int_0^\ell P_0^*P_0'\,dx}
=\frac{2}{r_0N}\int_0^\ell P_0^*B_0s(x)\,dx.}
$$
For either parity choice this expression is real. Indeed $|P_0|^2$, $A_0^2$ and $B_0^2$ are even about the midpoint, while $s$ is odd. Hence $\int|P_0|^2s=\int B_0^2s=0$, and the last numerator reduces to $\int A_0B_0s$.
No integral evaluation is needed to compare the two shifts. The phase change $P_Q=iP_D$ gives $A_Q=-B_D$, $B_Q=A_D$, and leaves $N$ and $I$ unchanged. Thus $A_QB_Q=-A_DB_D$, and
$$
\boxed{r_{1,Q}=-r_{1,D}.}
$$
Equivalently, the term involving $|P_0|^2$ vanishes by parity, while $(P_Q^*)^2=-(P_D^*)^2$ reverses the remaining perturbation numerator. This is <parity splitting of a mean-field dynamo threshold>: the perturbation breaks the phase degeneracy and lowers one onset threshold by the amount it raises the other.
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