Hmmm, an old question without a satisfactory answer. I'll have a go.

The spins of the two $B$ may combine as \begin{align} \text{singlet}\quad|s_1s_2\rangle &= \frac{\left|\uparrow\downarrow\right> - \left|\uparrow\downarrow\right\rangle}{\sqrt2}, & \text{or triplet}\quad|s_1s_2\rangle &= \frac{\left|\uparrow\downarrow\right> + \left|\uparrow\downarrow\right\rangle}{\sqrt2}. \end{align}

Since the two $B$ have spin-1/2, they obey Fermi-Dirac statistics and their total wavefunction must be antisymmetric under exchange. Therefore the antisymmetric spin singlet can only be paired with $L=0,2,\cdots$, which have even parity; the symmetric spin triplet must be paired with $L=1,3,\cdots$, so that the parity of the orbital angular momentum wavefunction makes the entire wavefunction change sign if the two $B$ are interchanged.

The intrinsic parity of $B$ doesn't contribute to the overall parity of the final state, since there are two of them; if the $B$ have negative parity, the overall intrinsic parity of the pair is still positive. Thus the allowed final states are \begin{align} \text{positive parity}&: & \text{spin singlet (antisymmetric)} && L&=0 \\ \text{negative parity}&: & \text{spin triplet (symmetric)} && L&=1 \end{align} If $A$ has definite parity and parity is conserved in the decay, only one of these possibilities is allowed.

I've omitted the spin singlet combined with $L=2$, since we do not have that much angular momentum. Likewise there is no combination of the spin triplet and $L=3$ which can be produced from a $J=2$ initial state.

Hmmm, an old question without a satisfactory answer. I'll have a go.

The spins of the two $B$ may combine as \begin{align} \text{singlet}\quad|s_1s_2\rangle &= \frac{\left|\uparrow\downarrow\right> - \left|\uparrow\downarrow\right\rangle}{\sqrt2}, & \text{or triplet}\quad|s_1s_2\rangle &= \frac{\left|\uparrow\downarrow\right> + \left|\uparrow\downarrow\right\rangle}{\sqrt2}. \end{align}

Since the two $B$ have spin-1/2, they obey Fermi-Dirac statistics and their total wavefunction must be antisymmetric under exchange. Therefore the antisymmetric spin singlet can only be paired with $L=0,2,\cdots$, which have even parity; the symmetric spin triplet must be paired with $L=1,3,\cdots$, so that the parity of the orbital angular momentum wavefunction makes the entire wavefunction change sign if the two $B$ are interchanged.

The intrinsic parity of $B$ doesn't contribute to the overall parity of the final state, since there are two of them; if the $B$ have negative parity, the overall intrinsic parity of the pair is still positive. Thus the allowed final states are \begin{align} \text{positive parity}&: & \text{spin singlet (antisymmetric)} && L&=0 \\ \text{negative parity}&: & \text{spin triplet (symmetric)} && L&=1 \end{align} I've omitted the spin singlet combined with $L=2$, since we do not have that much angular momentum. Likewise there is no combination of the spin triplet and $L=3$ which can be produced from a $J=2$ initial state.

If $A$ has definite parity and parity is conserved in the decay, only one of these possibilities is allowed. In fact, the way that angular momentum combines (with the possibility of the spin triplet plus the $L=1$ wavefunction adding to a $J^P=0^-$ final state) means that the parity has more to do with the allowed final spin state than $A$'s spin does: $$ \begin{array}{r|cc} & \text{spin 0} & \text{spin 1} \\ \hline \text{parity}+ & \text{decay to singlet},0^+ & \text{decay forbidden} \\ \text{parity}- & \text{decay to triplet},0^- & \text{decay to triplet},1^- \end{array} $$