PKCS Padding Method (Example 1)

Clear text consists of the following18 bytes:
F14ADBDA019D6DB7 EFD91546E3FF8444 9BCB
In order to make this a multiple of 16 bytes (the AES block size), we must add 14 bytes. Each byte will contain the value 0x0E, which is 14, the total number of padding bytes added. The result is that the padded clear text is as follows:
F14ADBDA019D6DB7 EFD91546E3FF8444 9BCB0E0E0E0E0E0E 
0E0E0E0E0E0E0E0E
The padded value is 32 bytes in length, which is two AES blocks. This padded string is encrypted in CBC mode, and the resulting ciphertext will also be 32 bytes in length.