I know that a perfectly random unbiased and not reused OTP cipher is perfect. An OTP under cryptanalysis has a very strong bias (up to 30 % of symbols are represented by 5 out of 255 possible symbols) but no repetition could be found. This OTP encryption is theoretically vulnerable. May I know the possible sequence of steps to exploite the vulnerability?
Assuming that there are no other flaws in the scheme, e.g. that the pads are not reused, the bias alone may or may not be practically exploitable. Whether or not it is mostly depends on how much you know (or can guess) about the plaintext.
For example, let's say you have a fairly short list of candidate plaintexts that the message might encode. Given each possible plaintext and the known ciphertext, you can now determine what the pad that yields that ciphertext from that plaintext would have to be, and then calculate the probability that the known biased pad generation process would produce that pad. This will then tell you which of the plaintexts is most likely to be the correct one (or, more precisely, let you update your prior estimates of the likelihood of each plaintext according to Bayes' theorem).
More generally, given sufficient bias in the pad and sufficient redundancy in the plaintext, you may be able to simply guess which characters of the ciphertext are likely to be encrypted with the most common pad values, and then fill in the missing characters. For example, the ciphertext below encodes a common English pangram, encrypted with a standard mod 26 one-time pad, biased so that a bit over 50% of all the pad characters are
A (effectively leaving about half of the plaintext characters unchanged).
Can you guess what the plaintext is?