= Solution
Interpret the multiplicative <Gaussian white noise> as the limit of a smooth short-correlation-time velocity field, hence in the <Stratonovich integral> convention. Put $\partial_i=\partial/\partial B_i$. For a realization, differentiating the field-space <Dirac delta> gives its field-space <continuity equation>
$$
\partial_t\widetilde P=-\partial_i(\sigma_{im}B_m\widetilde P).
$$
The velocity-gradient covariance has zero trace variance: contracting $T_{mn}^{ij}$ with $m=i,n=j$ gives $3-(9+3)/4=0$. Thus the gradient is trace free, consistently with the incompressible ideal induction equation.
A noise impulse in component $\sigma_{jn}$ changes $\widetilde B_j$ by $\widetilde B_n$ times that impulse. Its instantaneous retarded action on the <Dirac delta> is therefore
$$
\frac{\delta\widetilde P}{\delta\sigma_{jn}}=-\partial_j(B_n\widetilde P).
$$
Only earlier times affect the present field. In the white-noise limit the integral over this causal interval has $\int_0^t\delta(t-t')\,dt'=1/2$, corresponding to the symmetric short-time regularization. Applying the <Furutsu–Novikov formula> and then averaging gives
$$
\langle\sigma_{im}\widetilde P\rangle=-\frac{\kappa_2}{2}T_{mn}^{ij}\partial_j(B_nP).
$$
Substitute into the averaged field-space <continuity equation>:
$$
\partial_tP=\frac{\kappa_2}{2}\partial_i\left[B_mT_{mn}^{ij}\partial_j(B_nP)\right].
$$
The differentiated $B_n$ term vanishes because $\sum_jT_{mj}^{ij}=0$, and the remaining contraction is
$$
T_{mn}^{ij}B_mB_n=B^2\delta_{ij}-\frac12B_iB_j.
$$
Consequently the closed vector <Fokker-Planck equation> for <Gaussian white-noise magnetic stretching> is
$$
\boxed{\partial_tP=\frac{\kappa_2}{2}\partial_i\left[\left(B^2\delta_{ij}-\frac12B_iB_j\right)\partial_jP\right].}
$$
This is a positive <diffusion> operator: the displayed matrix has eigenvalues $B^2$ on the two transverse directions and $B^2/2$ along $\mathbf B$. Replacing the endpoint half-weight by one would double the <diffusion> and fail to reproduce the later rate.
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