Like, if we add 0999(i.e, 0000 1001 1001 1001) which has 3 digits and a sign digit to 0999(itself), we should get 01998 which cannot be represented with 3 digits and a sign bit.

I have an answer, let me know if you think the same. Overflow is detected in BCD addition, if the left-most digit i.e, the sign digit is not 0 or 9 after completion of the addition.