How should DES plaintext IP substitution be carried out

Hello, I've been reading one of your articles called

DES encryption and decryption(detailed encryption process)

One of the passages is

""
For example, on a block of M, perform the initial transform and get:

M = 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111
IP = 1100 1100 0000 0000 1100 1100 1111 1111 1111 0000 1010 1010 1111 0000 1010 1010
'''

I would like to ask how this paragraph came about. According to the PC-1 of the IP replacement table you gave in your article, 57,49,41 and so on, the third bit of IP should be 1 to

part or you have other circumstances(maybe you did not use the original PC-1 table here?).

Thanks very much for your time!!