How to do 2 variable XOR
Using De Morgan's laws and this question and following answers which contain some simplification tables :
A^B = A`B + AB`

(AC ^ BD) + A = (AC)`(BD) + (AC)(BD)` +A = (A`+C`)(BD) + (AC)(B`+D`) + A = A`BD + C`BD + AB`C + ACD` + A = A(B`C + CD` + 1) + A`BD + C`BD = A + A`BD + C`BD = A + BD + C`BD = A + BD(C` + 1) = A + BD

for learning purposes, if you want to get some clues without going into calculations, it's obvious that A is a part of answer because it's ORed with other stuff. and when it is, the C is somehow out of it because if the A=0 then the C vanishes and the BD comes out of XOR. if the A=1, the output is always 1 beacuse of ORed A. so the solution only depends on BD (first case) and A.