I expand everything:
$$\frac{1}{4}tr[\gamma^\alpha p_\alpha\prime \gamma^\mu (1+\gamma^5) \gamma^\beta p_\beta \gamma^\nu(1+\gamma^5)]$$
$$=\frac{1}{4}tr[\gamma^\alpha p_\alpha \prime \gamma^\mu\gamma^\beta p_\beta\gamma^\nu + \gamma^\alpha p_\alpha \prime \gamma^\mu \gamma^5 \gamma^\beta p_\beta\gamma^\nu+\gamma^\alpha p_\alpha \prime \gamma^\mu\gamma^\beta p_\beta\gamma^\nu\gamma^5+\gamma^\alpha p_\alpha \prime \gamma^\mu\gamma^5\gamma^\beta p_\beta\gamma^\nu\gamma^5]$$
swap $\gamma^5$ in second term for two times gets third term, swap $\gamma^5$ in forth term and using $\gamma^5\cdot\gamma^5=1$, it is the first term
$$=\frac{1}{4}tr[\gamma^\alpha p_\alpha \prime \gamma^\mu\gamma^\beta p_\beta\gamma^\nu + \gamma^\alpha p_\alpha \prime \gamma^\mu \gamma^\beta p_\beta\gamma^\nu \gamma^5+\gamma^\alpha p_\alpha \prime \gamma^\mu\gamma^\beta p_\beta\gamma^\nu\gamma^5+\gamma^\alpha p_\alpha \prime \gamma^\mu\gamma^\beta p_\beta\gamma^\nu\gamma^5\gamma^5]$$
$$=\frac{1}{4}tr[\gamma^\alpha p_\alpha \prime \gamma^\mu\gamma^\beta p_\beta\gamma^\nu + \gamma^\alpha p_\alpha \prime \gamma^\mu \gamma^\beta p_\beta\gamma^\nu \gamma^5+\gamma^\alpha p_\alpha \prime \gamma^\mu\gamma^\beta p_\beta\gamma^\nu\gamma^5+\gamma^\alpha p_\alpha \prime \gamma^\mu\gamma^\beta p_\beta\gamma^\nu]$$
$$=\frac{1}{2}tr[\gamma^\alpha p_\alpha \prime \gamma^\mu\gamma^\beta p_\beta\gamma^\nu + \gamma^\alpha p_\alpha \prime \gamma^\mu \gamma^\beta p_\beta\gamma^\nu \gamma^5]$$
$$=2(p\prime^\mu p^\nu+p\prime^\nu p^\mu-g^{\mu\nu}p\prime.p-i\epsilon^{\alpha\mu\beta\nu}p\prime_\alpha p_\beta)$$
For $|M|^2$, $2(p\prime^\mu p^\nu+p\prime^\nu p^\mu-g^{\mu\nu}p\prime.p-i\epsilon^{\alpha\mu\beta\nu}p\prime_\alpha p_\beta).2(k\prime_\mu k_\nu+k_\nu\prime k_\mu-g_{\mu\nu}k\prime.k-i\epsilon_{\rho\mu\sigma\nu}k^\rho k\prime ^\sigma)$
Terms like
$$p^\mu p\prime^\nu k_\mu k\prime_\nu = (p.k)(p\prime.k\prime)$$
$$p^\mu p\prime^\nu g_{\mu\nu}k\prime.k = (p.p\prime) (k\prime.k)$$
$$p^\mu p\prime^\nu i\epsilon_{\rho\mu\sigma\nu}k^\rho k\prime ^\sigma = i\epsilon_{\rho\mu\sigma\nu} p^\mu p\prime^\nu k^\rho k\prime ^\sigma$$
You got four terms like these but they cancel pairwise as,
$$i\epsilon_{\rho\mu\sigma\nu} p^\mu p\prime^\nu k^\rho k\prime ^\sigma+i\epsilon_{\rho\mu\sigma\nu} p^\nu p\prime^\mu k^\rho k\prime ^\sigma$$
$=i\epsilon_{\rho\nu\sigma\mu} p^\nu p\prime^\mu k^\rho k\prime ^\sigma+i\epsilon_{\rho\mu\sigma\nu} p^\nu p\prime^\mu k^\rho k\prime ^\sigma$ interchange dummy variable in first term
$=-i\epsilon_{\rho\mu\sigma\nu} p^\nu p\prime^\mu k^\rho k\prime ^\sigma+i\epsilon_{\rho\mu\sigma\nu} p^\nu p\prime^\mu k^\rho k\prime ^\sigma =0 $ swap index of $\epsilon$ in first term
$g^{\mu\nu}p\prime.p g_{\mu\nu}k\prime.k=4(p\prime .p)(k\prime .k)$ as summation over index we got a four
$g^{\mu\nu}p\prime.p i\epsilon_{\rho\mu\sigma\nu}k^\rho k\prime ^\sigma =0$ as $g^{\mu\nu}$ is non-zero only for repeated index but $\epsilon_{\rho\mu\sigma\nu}$ is zero for repeated index.
$$\epsilon^{\alpha\mu\beta\nu}p\prime_\alpha p_\beta
\epsilon_{\rho\mu\sigma\nu}k^\rho k\prime ^\sigma=\epsilon^{\nu\mu\beta\alpha}p\prime_\alpha p_\beta
\epsilon_{\nu\mu\sigma\rho}k^\rho k\prime ^\sigma=-2(\delta^\beta_\sigma \delta^\alpha_\rho-\delta^\beta_\rho \delta^\alpha_\sigma) p\prime_\alpha p_\beta k^\rho k\prime^\sigma =-2 (p.k\prime)(p\prime.k)+2(p\prime.k\prime)(p.k)$$
You can collect term by term and got PS results.
