= Solution
Every interaction contains an even number of $b$ <fields>, so the exact discrete <symmetry> $b\mapsto-b$ forbids $\Phi\to bbb$ to every perturbative order. There is no such prohibition for $\Phi\to bb$.
The latter <decay channel> is absent at <tree-level Feynman diagram>[tree level] in the displayed interactions, but contracting two $b$ legs at a $\Phi b^2(\partial b)^2$ vertex produces a <tadpole diagram> with two external $b$ legs. For example, contracting the undifferentiated pair leaves the local operator $\Phi(\partial b)^2$ multiplied by an integral proportional to $\int d^4\ell/(\ell^2+i0)$. Contracting the differentiated pair instead gives a local $\Phi b^2$ contribution. Both are allowed <effective field theory> operators. With a <ultraviolet cutoff>, these local terms generically carry quadratic or quartic cutoff dependence and require matching coefficients or <counterterms>.
There is a relevant qualification to a claim that the one-loop <decay width> must be nonzero. Since $b$ is exactly massless here, the one-vertex <tadpole integrals> have no external scale and are <scaleless integrals>. They vanish in <dimensional regularization>. Thus the correct conclusion is \b[allowed, and generically generated by quantum corrections or matching], not a regulator-independent nonzero one-loop prediction from $\lambda_b$ alone. No exact <symmetry> protects the two-$b$ channel; its numerical rate requires the renormalized coefficients of the allowed operators.
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