Padding is incorrect. AES encryption - encryption

Question

encrypted=b'$\x940\xc6\xef3s\x01|e\xe8\xdf\xaa00\xcc\x8c\x98i\x84\xcc[e\xd2\xa4]\xb37\xab\xbc#} \xaeC\xdd\xd7\x89\xdd[\xab)\xf0t\xa8\x95\xfe\xf9>\x15\x02:\x85-s*\x0cf\xe1\xa5\xc3\xa8]\x1d\x8f\xe2f\x1aje\x9f\xc3^\xe0\xc6i\xd6\x02$I\xce0_\x97\x8c\x06Rj\x1e\xe0F\x1fE\xfb\xf4\x0be:\x01\xd1m\xe4\xfaVv\xa0\xdf\xa8\x1e\xec\x03\xc1\x9a\x89\x9a\xa1m\xb4uL\x97\x81\xdc\xaaG5\xf7Q\xba\xc6\xf33\x05\x94l#pL!\x886\x9a5\xa3\xa2Z)\xd3\x80&p\xa4)\x1e\x9ei\xcc\xa9\xcc\xe9j\xe9\xb4\x14\xe1<\x8c\xa50\xd4\xf9\xb5\xb2\xd4#3\nwW#m\x14\xb67\xf3!a\x19:\x8af\x1c\xf1\x13\xbfM\x1a\xfe\xaf\xb0\x95Y~\xe2}\x06\xf0\xc1\xff\xde\x04\x06\xc9\xb4\xfe\xc8\xbb\x8eS\xff\xe29Q\xa0\xf6\xd2\xc78\xf5\x97\x07OW\xbb\xfd\x08q\x8f4C\xbb\x00\xa0\xa1_g\x0e3&x\\\x1a\xa4/a\xa2\x1bp|\x84\xdfr[.\xe5\xbd\xb1\xfc\xc3\t_0l\xddU\xc8HL#&LD\x17g=\xb5\xb0\xcf'
mk123=b'$\x940\xc6\xef3s\x01|e\xe8\xdf\xaa00\xcc\x8c\x98i\x84\xcc[e\xd2\xa4]\xb37\xab\xbc#} \xaeC\xdd\xd7\x89\xdd[\xab)\xf0t\xa8\x95\xfe\xf9>\x15\x02:\x85-s*\x0cf\xe1\xa5\xc3\xa8]\x1d\x8f\xe2f\x1aje\x9f\xc3^\xe0\xc6i\xd6\x02$I\xce0_\x97\x8c\x06Rj\x1e\xe0F\x1fE\xfb\xf4\x0be:\x01\xd1m\xe4\xfaVv\xa0\xdf\xa8\x1e\xec\x03\xc1\x9a\x89\x9a\xa1m\xb4uL\x97\x81\xdc\xaaG5\xf7Q\xba\xc6\xf33\x05\x94l#pL!\x886\x9a5\xa3\xa2Z)\xd3\x80&p\xa4)\x1e\x9ek\xcc\xa9\xcc\xe9j\xe9\xb4\x14\xe1<\x8c\xa50\xd4\xf9\xb5\xb2\xd4#3\nwW#m\x14\xb67\xf3!a\x19:\x8af\x1cq\x13\xbfM\x1a\xfe\x8f\xb0\x95Y~\xe2}\x06\xf0\xc1\xff\xde\x14\x06\xc9\xb4\xfe\xc8\xb3\x8eS\xff\xe29Q\xa0\xf6\xd2\xc78\xf5\x97\x07OW\xbb\xfd\x08q\x8f4C\xbb\x00\xa0\xa1_g\x0es&x\\\x1a\xa4/a\xa2\x1bp|\x84\xdbr[.\xe5\xbd\xb1\xbc\xe3\t_0L\xd9U\xc8HL#&LD\x17g=\xb5\xb0\xcf'
note:ive done some process in between so the bytestrings mk123 and encrypted are not the same
the first byte string is the one ive got after encryption,but when i try to decode its throwing the error"padding is incoreect"
this is the code ive used
from hashlib import md5
from Crypto.Cipher import AES
from Crypto.Random import get_random_bytes
from Crypto.Util.Padding import pad, unpad
class AESCipher:
def __init__(self, key):
password = key.encode('utf-8')
self.key = md5(password).digest()
def encrypt(self, data):
vector = get_random_bytes(AES.block_size)
encryption_cipher = AES.new(self.key, AES.MODE_CBC, vector)
return vector + encryption_cipher.encrypt(pad(data, AES.block_size))
def decrypt(self, data):
file_vector = data[:AES.block_size]
decryption_cipher = AES.new(self.key, AES.MODE_CBC, file_vector)
return unpad(decryption_cipher.decrypt(data[AES.block_size:]), AES.block_size)
the driver code is
message='0000000000000000011111111111111001000010001011100101101011101110010000100010111001011010111011100101101000100010011111111111111001111111111111100101111011100010010111101110101001011110111010100101111011101010010001100010001001111111111111100000000000000000'
msg = message.encode('utf-8')
pwd = "password"
encrypted = AESCipher(pwd).encrypt(msg)
print('Ciphertext:', encrypted)
decrypted = AESCipher(pwd).decrypt(mk123).decode('utf-8')

Answer 1

If you change the last block or the previous to last block of ciphertext then unpadding may indeed fail for CBC mode. This is because both of these blocks change the resulting plaintext. Generally, for CBC mode, you'll change the bits of one block with a chance of 0.5 for the block that you changed, and the same bits of the plaintext at the next block are flipped. How changes affect the decryption routine is called "error propagation".
You can figure most of this out yourself by looking at the decryption diagram from Wikipedia (by WhiteTimberwolf):
Knowing that, if you want to investigate the original plaintext you simply need to skip unpadding. Then you get the above changes made to the plaintext (including padding). You can then try to fix or guess the plaintext and/or correct the padding. If you cannot correct the padding then the size of the plaintext can differ by one block - as far as the cipher is concerned anyway.