Using Python, write a 7 bit PDU encoder that accept plaintext

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Python

I was given a task in python to write a 7 bit PDU encoder that accepts a plaintext ascii string and returns a tuple (x,y)
where x is the encoded plaintext formatted as a binary sting with each byte separated with 5 hashes
‘#####’, and y is the hexadecimal representation of the string.

Am just confused about this question,


[EDIT: Moved up here from code below]
I don't know how to make the code to be separated with 5 hashes (#####) and how to make it display both ascii and binary. I appreciate any help from you.

What I have tried:

# -*- coding: utf8 -*-
import binascii
pdu = ("@£$¥èéùìòÇ\nØø\rÅåΔ_ΦΓΛΩΠΨΣΘΞ\x1bÆæßÉ !\"#¤%&'()*+,-./0123456789:;<=>?"
       "¡ABCDEFGHIJKLMNOPQRSTUVWXYZÄÖÑܧ¿abcdefghijklmnopqrstuvwxyzäöñüà")
ext = ("````````````````````^```````````````````{}`````\\````````````[~]`"
       "|````````````````````````````````````€``````````````````````````")

def pdu_encode(m):
    res = ""
    for i in m:
        idx = pdu.find(i);
        if idx != -1:
            res += chr(idx)
            continue
        idx = ext.find(i)
        if idx != -1:
            res += chr(27) + chr(idx)
    return binascii.b2a_hex(res.encode('utf-8'))

print(pdu_encode("Cryptomasses on groove"))