To find the perimeter of the figure, we need to sum the lengths of all its outer edges.

To find the perimeter of the figure, we need to sum the lengths of all its outer edges.

Let's identify the lengths of the exterior sides:

1. The left vertical side of rectangle B: $5 \text{ cm}$
2. The bottom horizontal side of rectangle B: $12 \text{ cm}$
3. The right vertical side of rectangle B: $5 \text{ cm}$
4. The top horizontal side of rectangle B that is exposed (not covered by A): $9 \text{ cm}$
5. The left vertical side of rectangle A: $3 \text{ cm}$
6. The top horizontal side of rectangle A: $3 \text{ cm}$
7. The right vertical side of rectangle A: $3 \text{ cm}$

Now, we sum these lengths to find the total perimeter:

$$P = 5 \text{ cm} + 12 \text{ cm} + 5 \text{ cm} + 9 \text{ cm} + 3 \text{ cm} + 3 \text{ cm} + 3 \text{ cm}$$ $$P = 40 \text{ cm}$$

Alternatively, we can calculate the perimeter of each rectangle and subtract the overlapping internal boundary.
Perimeter of rectangle B: $2 \times (12 \text{ cm} + 5 \text{ cm}) = 2 \times 17 \text{ cm} = 34 \text{ cm}$
Perimeter of rectangle A: $2 \times (3 \text{ cm} + 3 \text{ cm}) = 2 \times 6 \text{ cm} = 12 \text{ cm}$
The length of the common boundary (where A and B meet) is $3 \text{ cm}$. This boundary is counted twice when summing the individual perimeters, so we subtract $2 \times 3 \text{ cm}$ from the sum.
Total perimeter = (Perimeter of B) + (Perimeter of A) - $2 \times (\text{common boundary})$

$$P = 34 \text{ cm} + 12 \text{ cm} - 2 \times 3 \text{ cm}$$ $$P = 46 \text{ cm} - 6 \text{ cm}$$ $$P = 40 \text{ cm}$$

Both methods yield the same result.

Comparing with the given options:

(a) $69 \text{ cm}$

(b) $46 \text{ cm}$

(c) $43 \text{ cm}$

(d) $40 \text{ cm}$

(e) $37 \text{ cm}$

The calculated perimeter matches option (d).

The final answer is $\boxed{\text{40cm}}$.