I am unsure what you mean by "solving" this, but the meaning of the code is rather obvious and even more so in the screenshot you provided. Simplified version:
add [esp], 5
xor eax, eax
And with annotations:
call $+5 ; call address of next instruction, placing return address on stack (esp)
add [esp], 5 ; add 5 bytes to the return address
ret ; return to the address from the stack ... which happens to be
xor eax, eax ; ... this instruction
ret ; return for good from _main ...
So what this does is to return 0 as a 32-bit value from
Given the (decompiled) condition:
if ((cVar1 = func_0x8b4c55a0(), cVar1 == -0xe) || 1)
... this doesn't change a thing, however, as the
|| 1 will ensure that this condition is always true and
cVar1 doesn't subsequently get used.
So as far as you provide context, this is pure distraction.