Here is another solution:X is infinite and so there are two elemnts x1 and x2.Let X'=X{x1,x2}.I have that |X|=|X'|and so |P(X)|=|P(X')|.Let A be a subset of X' and so |X\A|>=2.Then there exist a permutation f:X\A----> X\A without fixed points.Then I extend f to X leaving fixed the elemnts of A.The set of the points fixed by f is then A.So I have a surjection Sym(X)---> P(X').Hence |Sym(X)|>=|P(X')|=|P(X)|.That |Sym(X)|<=|P(X)| is easy.

Here is another solution: $X$ is infinite and so there are two elements $x_1$ and $x_2$. Let $X'=X\setminus \lbrace x_1,x_2\rbrace$. I have that $|X|=|X'|$ and so $|P(X)|=|P(X')|$. Let $A$ be a subset of $X'$ and so $|X\setminus A|\ge 2$. Then there exist a permutation $f:X\setminus A \to X\setminus A$ without fixed points. Then I extend $f$ to $X$ leaving fixed the elements of $A$. The set of the points fixed by $f$ is then $A$. So I have a surjection $Sym(X) \to P(X')$. Hence $|Sym(X)| \ge |P(X')|=|P(X)|$. That $|Sym(X)| \le |P(X)|$ is easy.