= Solution
In the matrix expression, $t^a=\lambda^a/2$ denotes a fundamental generator, whereas the component <adjoint covariant derivative> uses adjoint generators. With $\operatorname{Tr}(t^dt^h)=\delta^{dh}/2$,
$$
[t^a,\Phi]=if^{acd}\phi_ct^d,\qquad
-g^2\operatorname{Tr}([t^a,\Phi][t^b,\Phi])B_\mu^aB^{b\mu}
=\frac{g^2}{2}f^{acd}f^{bed}\phi_c\phi_eB_\mu^aB^{b\mu}.
$$
Using antisymmetry to write $f^{acd}=f^{dac}$ and $f^{bed}=f^{dbe}$ gives exactly the component <mass term>. This is the <gauge-boson mass matrix from an adjoint scalar>.
For $\Phi_0=v\operatorname{diag}(1,1,-2)$, the <Gell-Mann matrices> $t^1,t^2,t^3,t^8$ commute with $\Phi_0$. Their <gauge bosons> are therefore massless. The generators $t^4,t^5,t^6,t^7$ connect the first two <eigenspaces> to the third; the <eigenvalue> difference is $3v$. Directly taking the <matrix trace> gives
$$
\operatorname{Tr}([t^a,\Phi_0][t^b,\Phi_0])=-\frac{9v^2}{2}\delta^{ab},\qquad a,b\in\{4,5,6,7\},
$$
with all remaining entries zero. Thus
$$
\mathcal L_{\rm mass}=\frac{9g^2v^2}{2}\sum_{a=4}^7 B_\mu^aB^{a\mu},\qquad
\boxed{m_{1,2,3,8}=0,\qquad m_{4,5,6,7}=3g|v|.}
$$
The factor $1/2$ is the canonical real-vector <mass term> normalization. The unbroken <Lie algebra> is $\mathfrak{su}(2)\oplus\mathfrak u(1)$, giving the usual description \b[SU(2) × U(1)]. More precisely, the <centralizer> of the <vacuum expectation value> inside $SU(3)$ is
$$
\boxed{S(U(2)\times U(1))\simeq (SU(2)\times U(1))/\mathbb Z_2.}
$$
This global quotient refines the <adjoint-Higgs breaking of SU(3) to SU(2) and U(1)> without changing the local <gauge boson> spectrum.
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