The only candidate key is $\{ABD\}.$
So, $AB \to C, BD \to EF$ and $A \to G$ violates 2NF. To make the relation in 2NF, we can decompose $R$ to $R_1(A,B,C), R_2(A,B,D),R_3(A,G), R_4(B,D,E,F,H)$
In the above decomposition we do not have any partial FD but transitive dependency $F \to H$ is there. To avoid this $R_4$ must be split to $R_5(B,D,E,F)$ and $R_6(F,H).$
Thus $a = 4, b = 5 \implies a \times b = 20.$