This is a Reverse Engineering Commodore challenge.
We use DirMaster to find the code of the program
2 gosub 1000
4 input b$
10 gosub 4000
13 poke 11,tt-47
30 lli = lli + 1
31 xd = asc(mid$(a$,lli,1))
35 if xd > 127 then xd = xd - 160
40 poke 251,xd
41 poke 252,dk
50 sys 49152
51 dk = peek(253)
52 poke 252,dk
53 poke 251,peek(11)
54 sys 49152
98 print chr$(peek(253));
99 if peek(6) <= lli then goto 3000
100 goto 30
1000 print "searching for a secret file on drive 8.."
1001 open 2,8,2,"secret,s,r"
1002 input#2,a$
1003 if st<>0 then print "cannot find a secret file" :close 2 :end
1004 print "secret file found!"
1005 print "enter decryption key (8 chars)"
1006 poke 6,len(a$) :
1008 lli = 0
1009 return
1500 rem
1501 poke 250,0
1505 poke 251,y
1506 sys 49159
1510 for x=1 to 6
1515 z = asc(mid$(b$,x,1))
1520 poke 251,z
1530 poke 252,y
1540 sys 49152
1546 sys 49159
1550 y = peek(253)
1560 next x
1600 return
2000 rem
2020 for i=0 to 17
2030 read a: poke 49152+i,a
2040 next
2060 restore
2070 return
3000 end
4000 rem
4001 close 2
4002 if len(b$) <> 8 then print "invalid key length!" : goto 3000
4018 gosub 2000
4020 y = asc(mid$(b$,7,1))
4025 gosub 1500
4026 dk = peek(250)
4030 if y <> asc(mid$(b$,8,1)) then print "invalid key!" : goto 3000
4049 read tt
4050 return
10000 data 165,251,69,252,133,253,96,6,250,165,251,41,1,5,250,133,250,96
Now, lets add some comments to the last part, which is the most important:
Section 2000
This section is as follows:
2000 rem
2020 for i=0 to 17
2030 read a: poke 49152+i,a
2040 next
2060 restore
2070 return
3000 end
4000 rem
This is a machine code loader. It loads the instructions from data.
In this case,
10000 data 165,251,69,252,133,253,96,6,250,165,251,41,1,5,250,133,250,96