It seems that they are correct. The simplest form of the problem can be thought of as the following
$$ \max_{\alpha, p} \; \alpha \log(1+p) \quad \text{s.t.} \quad \alpha p \le 1,\; \quad \alpha,p \ge0$$
and they do a change of variable $m = \alpha p$ to obtain
$$ \max_{\alpha, m} \; \alpha \log(1+m/\alpha) \quad \text{s.t.} \quad m \le 1, \quad \alpha, m \ge 0.$$
The constraints of this problem are clearly convex (i.e., affine inequalities). It only remains to verify that $f(\alpha,m) = \alpha \log(1+m/\alpha)$ is concave jointly in its variables. This is easy to verify using the Hessian
$$
\nabla^2 f(\alpha,m) = \frac{1}{(\alpha + m)^2}
\begin{pmatrix}
-m^2/\alpha & m \\ m & - \alpha
\end{pmatrix}.
$$
The matrix above has nonpositive diagonals over $C = \{(\alpha,m) \mid \alpha,m \ge 0\}$ and the determinant is $0$. Hence, $\nabla^2 f(\alpha,m) $ is nonpostive definite over $C$, i.e. $f$ is concave over $C$. Since maximizing a concave function over a convex set is a convex optimization problem, the second problem is convex.
Here is a plot of function $f$:
Extension. What they in fact have is more like
$$ \max_{\alpha, p, \beta, q} \; \min\{\alpha \log(1+p), \beta\log (1+q)\} \quad \text{s.t.} \quad \alpha p \le 1,\quad \beta q \le 1\quad \alpha,p, \beta,q \ge0$$
which they turn into
$$ \max_{\alpha, m, \beta, s} \; \min\{\alpha \log(1+m/\alpha), \beta\log (1+s/\beta)\} \quad \text{s.t.} \quad m \le 1,\quad s \le 1\quad \alpha,m, \beta,s \ge0.$$
The convexity of the second problem follows from the above argument and that the minimum of a bunch of concave functions is again concave.
Update. Here is another trick they are using. Suppose you want to solve the following problem,
$$
(P) \quad \max_x \sum_{k=1}^n \min(g_{1,k}(x),g_{2,k}(x)).
$$
Let $g_k(x) =\min(g_{1,k}(x),g_{2,k}(x))$. Assuming that $g_{1,k}$ and $g_{2,k}$ are concave functions, then so is $g_k$ (check!). So the above problem is a convex problem. We can write it in epigraph form:
$$
(P') \quad \max_{x, t_1,\dots,t_n} \; \sum_{k=1}^n t_k \quad \text{s.t.} \quad
t_k \le \min(g_{1,k}(x),g_{2,k}(x)) , \;\text{for all $k=1,\dots,n$}.
$$
To see that this is equivalent to the original problem, first optimize over $t_1,\dots,t_k$ (and then over $x$). Optimizing over $t_k$ forces us to have $t_k = \min(g_{1,k}(x),g_{2,k}(x))$, giving back the original problem $(P)$.
Now, we can equivalently write $(P')$ as
\begin{align*}
(P'') \quad \quad &\max_{x, t_1,\dots,t_n} \; \sum_{k=1}^n t_k \quad \text{s.t.} \\ &\qquad
t_k \le g_{1,k}(x), \quad, t_k \le g_{2,k}(x) , \;\text{for all $k=1,\dots,n$}.
\end{align*}
This problem is equivalent to the original problem. If $x^*,t_1^*,\dots,t_k^*$ is an optimal solution of $(P'')$, then $x^*$ is an optimal solution of $(P)$ and we have $t^*_k = \min(g_{1,k}(x^*),g_{2,k}(x^*)) = g_k(x^*)$
