= Solution
Reading the arrows in the original diagram gives parents $\operatorname{pa}(A)=\operatorname{pa}(B)=\varnothing$, $\operatorname{pa}(C)=\{A,B\}$ and $\operatorname{pa}(D)=\{B\}$. The <Bayesian network> factorization is
$$
\boxed{P(a,b,c,d)=P(a)P(b)P(c\mid a,b)P(d\mid b).}
$$
For binary variables, each <conditional probability> table has one free <probability> per parent configuration, since its two entries sum to one. There are one each for $A$ and $B$, four for $C$, and two for $D$, giving
$$
\boxed{1+1+4+2=8\text{ free parameters}.}
$$
This is the dimension of the unrestricted binary <Bayesian network> family; extra <statistical parameter> equalities or deterministic relationships would describe a smaller model and are not imposed by the diagram.
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