Suppose the finite group $G$ has an even number of elements. (Note: clearly $4p$ is even.)
Imagine a Cayley table (a group multiplication table).
Suppose for the sake of contradiction that no element has order $2$.
Then there is only one $e$ along the NW/SE diagonal of the table; namely, for $e*e = e$.
But we know that $xy = e$ implies $yx = e$ as well, so every other $e$ in the table comes in a pair symmetric around the aforementioned diagonal.
Therefore, the number of $e$'s in the table is odd: the one in the diagonal and the rest in pairs.
But the number of $e$'s in the table is also even: one in each row, and $|G|$ is even.
Thus, we were wrong in our supposition, and there must be an element of order $2$.