If $q=p^a$ , where $p$ is a prime number, then I would like to know the number of conjugacy classes related to elements of order $p$ and $2$ in the simple group $PSL(2,q)$ . 



The number of conjugacy classes of the simple group PSL(2,q)
There is one conjugacy class of elements of order 2, and if $p$ is odd then there are two conjugacy classes of elements of order $p$. This goes back to Dickson's 1901 book on Linear Groups.


Added later: for order $p$ elements this can be seen as follows. All order-$p$ elements of $PSL(2,q)$ are conjugate to elements of any prescribed Sylow $p$-subgroup. One such Sylow $p$-subgroup $S$ of $PSL(2,q)$ consists of the upper-triangular matrices with $1$'s on the diagonal. Now consider the action of $PSL(2,q)$ on the projective line $\mathbb{P}^1(\mathbb{F}\_q)$. Each nonidentity element of $S$ has a unique fixed point, namely $\infty$. Thus, any element of $PSL(2,q)$ which conjugates one nonidentity element of $S$ to another must fix $\infty$, and hence must be upper-triangular. Finally, one easily checks that
$$
\left(\begin{matrix} a & b \\ 0 & a^{-1} \end{matrix}\right)
\left(\begin{matrix} 1 & c \\ 0 & 1 \end{matrix}\right)
\left(\begin{matrix} a & b \\ 0 & a^{-1} \end{matrix}\right)^{-1}
=
\left(\begin{matrix} 1 & a^2 c \\ 0 & 1 \end{matrix}\right).
$$
It follows that the conjugacy classes of order-$p$ elements are in bijection with $\mathbb{F}\_p^\times/(\mathbb{F}\_p^\times)^2$, so there are two such classes if $p$ is odd and one if $p$ is even.


For order $2$ I don't know a proof from first principles that is as short as the one above; the quickest proof I know is the one given in the proof of Lemma A.3 of my paper with Bob Guralnick titled "Polynomials with PSL(2) monodromy".


