We need to invite a total of $8$ people such that:

• $4$ are boys and $4$ are girls
• $5$ are from group $A$
• $3$ are from group $B$

Group composition:

• Group $A$: $7$ boys, $3$ girls
• Group $B$: $6$ boys, $5$ girls

Let the number chosen from group $A$ be:

• $x$ boys and $y$ girls Then: $x+y=5$

Since total invited are $4$ boys and $4$ girls, the number chosen from group $B$ will be:

• $4-x$ boys and $4-y$ girls

Also, from group $B$, total chosen is $3$: $(4-x) + (4-y) = 3$ $8-(x+y)=3$ $8-5=3$ which is consistent.

Now find all valid $(x,y)$ such that selections are possible.

Since group $A$ has only $3$ girls, we must have $y \le 3$. Also total girls invited are $4$, so $4-y \le 5$ in group $B$, always okay. Since total boys invited are $4$, $x \le 4$.

From $x+y=5$, possible cases are:

• $(x,y)=(2,3)$
• $(x,y)=(3,2)$
• $(x,y)=(4,1)$

Case $(5,0)$ is not possible because then group $B$ would need $-1$ boys? Actually from total boys/girls it gives group $B$: $(4-5,4-0)=(-1,4)$, impossible. Case $(1,4)$ impossible since group $A$ has only $3$ girls.

Count each case.

Case 1: Group $A$ contributes $2$ boys and $3$ girls

Then group $B$ contributes $2$ boys and $1$ girl.

Number of ways: $\binom{7}{2}\binom{3}{3}\binom{6}{2}\binom{5}{1}$ $=21 \cdot 1 \cdot 15 \cdot 5 = 1575$

Case 2: Group $A$ contributes $3$ boys and $2$ girls

Then group $B$ contributes $1$ boy and $2$ girls.

Number of ways: $\binom{7}{3}\binom{3}{2}\binom{6}{1}\binom{5}{2}$ $=35 \cdot 3 \cdot 6 \cdot 10 = 6300$

Case 3: Group $A$ contributes $4$ boys and $1$ girl

Then group $B$ contributes $0$ boys and $3$ girls.

Number of ways: $\binom{7}{4}\binom{3}{1}\binom{6}{0}\binom{5}{3}$ $=35 \cdot 3 \cdot 1 \cdot 10 = 1050$

Total number of ways: $1575+6300+1050=8925$

Therefore, the correct option is: $\boxed{8925}$ which is Option A.