Running file command on the file shows the file to be an ELF 32 bit LSB executable:

$ file chal
chal: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked, interpreter /lib/ld-linux.so.2, not stripped

Running strings on the file provides a huge chunk of data, while passing through head chal prints the displays

              8&6FVfv���/flag.txtnope
Got input: '%s'
Enter the password: e1n5tds4bu4et1n4ui2oh2ou4s4tu2oerror 

This text after password particularly stood out, and initating suffering to connect to the machine through netcat request password, which points out the text above to part of the password.

$ nc kashictf.iitbhucybersec.in 19787
Enter the password:

Upon opening the file within ghidra, and viewing the decompiled main, it can be seen that the text above goes through a number of transformation

  *(char **)(&sel_data)[on] = s_e1n5tds4bu4et1n4ui2oh2ou4s4tu2o_08050076;
  stack_temp = s_Enter_the_password:_08050061;
  *(undefined4 *)(&sel_data)[on] = *(undefined4 *)(sp + -0x200068);
  *(char **)(&sel_data)[on] = stack_temp;
  stack_temp = (char *)CONCAT22((short)*(undefined4 *)
                                        ((&alu_add16)
                                         [*(int *)(PTR_alu_add16_34820_08085fb0 + 0x20000)] +
                                        (*(uint *)(&alu_add16)[*(int *)PTR_alu_add16_48410_08093408 ]
                                        >> 0x10) * 4),
                                (short)*(uint *)(&alu_add16)[*(int *)PTR_alu_add16_48410_08093408] );
  *(undefined4 *)(&sel_data)[on] = *(undefined4 *)(sp + -0x200068);
  *(char **)(&sel_data)[on] = stack_temp;
  *(undefined4 *)(&sel_data)[on] = *(undefined4 *)(sp + -0x200060);
  *(undefined4 *)(&sel_data)[on] = *(undefined4 *)(sp + -0x200060);
  stack_temp = *(char **)(&sel_data)[on];

To avoid the need to fully understand the transformation, I passed the text within cyberchef and attempted a number of recipes. Passing through different base values, and ROT13 bruteforce, until coming across ROT47 bruteforce with the value 93, the text stood out as seenable below:

Amount = 91: p\b.k2qap1_r1bq.k1rf/le/lr1p1qr/l\-5-2--43
Amount = 92: q]c/l3rbq2`s2cr/l2sg0mf0ms2q2rs0m].6.3..54
Amount = 93: r^d0m4scr3at3ds0m3th1ng1nt3r3st1n^/7/4//65

As seen from the text, the d0m4scr3at3ds0m3th1ng1nt3r3st1n is sensible as a password. Passing the text as a password ultimately captured the flag.

$ nc kashictf.iitbhucybersec.in 19787
Enter the password: d0m4scr3at3ds0m3th1ng1nt3r3st1n
d0m4scr3at3ds0m3th1ng1nt3r3st1n
Got input: 'd0m4scr3at3ds0m3th1ng1nt3r3st1n'
KashiCTF{d1d_y0u_r3v_17_LByDG02ak}

It can be concluded that the transformation was a ROT47 with a value of 47.