Given an integer $N > a$, we claim that there exist an integer $d$ and a prime $p$, both greater than $N$, such that $d$ divides $ap+1$, $d$ and $(ap+1)/d$ are coprime, and $\sigma(d)/d > \sigma(a)$. In this case,
$$\sigma(ap + 1) = \sigma\left(\frac{ap+1}{d} \cdot d\right) = \sigma\left(\frac{ap+1}{d}\right) \sigma(d) > \frac{ap+1}{d} \cdot \sigma(d) \\ > (p+1)\sigma(a) = \sigma(p)\sigma(a) = \sigma(ap),$$
and we are done. Back to the claim, let $p_i$ be the $i$-th prime greater than $N$, take $k$ large enough so that $\sum_{i=1}^k 1/p_i > \sigma(a)$ – this is possible, for $\sum_{q \text{ prime}} 1/q = \infty$ – and set $d = p_1p_2\cdots p_k$. Then
$$\sigma(d)/d = \prod_{i=1}^{k} \left(1 + \frac{1}{p_i}\right) > 1 + \sum_{i=1}^{k} \frac{1}{p_i} > \sigma(a).$$
Next, use the Chinese remainder theorem to produce an integer $t$, which is unique modulo $p_1^2 p_2 \cdots p_k^2$, such that $at + 1 \equiv p_i \pmod{p_i^2}$, $i = 1, 2, \dots, k$; this is possible, for each $p_i > N > a$. Finally, use Dirichlet's theorem to pick up a prime $p > N$ from the arithmetic sequence
$$t + rp_1^2 p_2^2 \cdots p_k^2, \quad r = 0, 1, 2, \dots$$
Clearly, such a $p$ satisfies the stated conditions.