= Solution
We take $\alpha,\beta>0$, as is necessary for the assertion. The nonvacuous cases have $0<\alpha\leq1/2$ and $0<\beta\leq1$. Put $n=|A|$ and
$$
K=\max_{b\in B}\frac{|A+bA|}{n}.
$$
We prove a lower bound for $K$ by transferring small <sumset> growth through algebraic operations on the <scalars>. This gives a self-contained <polynomial skew-sumset expansion over a prime field> proof using the permitted <Ruzsa triangle inequality>, <Ruzsa covering lemma> and <Plünnecke-Ruzsa inequality>.
Since $|B|>1$, choose $b_0\in B\setminus\{0\}$. The bound $|b_0A+A|\leq K|b_0A|$ and the mixed-set <Plünnecke-Ruzsa inequality> give
$$
|rA-sA|\leq K^{r+s}n\qquad(r,s\text{ nonnegative integers}).
$$
In particular $|A+A|\leq K^2n$ and $|3A-2A|\leq K^5n$. Here repeated-set sums such as $3A$ allow independent choices from each copy.
For a <scalar> $t$, write $D(t)=|A+tA|/n$, and put $Q=A-A$ for the <difference set>. The <Ruzsa covering lemma> gives $tA\subseteq T_t+Q$ with $|T_t|\leq D(t)$. Indeed choose a maximal family of disjoint translates $u+A$, with $u\in tA$; counting their union bounds the family size, and maximality forces every remaining translate to meet one of them.
For any $r,s$, the containments $A+(r+s)A\subseteq A+rA+sA$ and $A+rsA\subseteq rT_s+A+rQ$ give, after covering each copy of $rA,sA$,
$$
D(r+s)\leq K^5D(r)D(s),\qquad D(rs)\leq K^5D(r)^2D(s).
$$
For the second bound, specifically $rQ=rA-rA\subseteq T_r-T_r+2Q$, and $A+2Q=3A-2A$. The <Ruzsa triangle inequality> also gives $|A-rA|\leq|A+A|\,|A+rA|/n$, so $D(-r)\leq K^2D(r)$. These are the <closure bounds for small skew-sumset scalars>.
We next prove a growth lemma for the <scalar> sets. For $S\subseteq\mathbb F_p$ with $m=|S|\geq2$, let
$$
R=\left\{\frac{a-b}{c-d}:a,b,c,d\in S,\ c\ne d\right\}.
$$
If $R\ne\mathbb F_p$, then $R+1$ is not contained in $R$: otherwise equal <cardinality> would make $R$ invariant under addition of one, which generates the entire <prime field>. Choose $r=1+(a-b)/(c-d)\notin R$. The map $(x,y)\mapsto x+ry$ on $S^2$ is injective. Indeed a collision with different second coordinates would express $r$ as a ratio of two differences from $S$, while equal second coordinates force equal first coordinates. Clearing the nonzero denominator puts its image inside
$$
(c-d)S+(c-d+a-b)S\subseteq3SS-3SS,
$$
so the latter set has at least $m^2$ elements. Here $SS$ is the <product set>, and $3SS-3SS$ denotes three independent product terms minus three others, not multiplication by the <scalar> three.
If $R=\mathbb F_p$, let $E_t$ count the collisions $x_1+ty_1=x_2+ty_2$ with all four variables in $S$. Summing over $t$ gives
$$
\sum_{t\in\mathbb F_p}E_t=pm^2+m^3(m-1)\leq pm^2+m^4.
$$
The first term comes from equal second coordinates, which force equal first coordinates. For unequal second coordinates, each quadruple determines exactly one <scalar>. Some $t$ consequently satisfies $E_t\leq m^2+m^4/p$. The <Cauchy-Schwarz inequality> on the representation counts gives
$$
|S+tS|\geq\frac{m^4}{E_t}\geq\frac{m^2}{1+m^2/p}\geq\tfrac12\min(p,m^2).
$$
Representing $t$ as a ratio from $R$ and clearing its denominator embeds a dilate of this <sumset> in $2SS-2SS\subseteq3SS-3SS$, padding with the same product in the positive and negative sums. We have proved the <quadratic growth of a sixfold product difference set> estimate
$$
\boxed{|3SS-3SS|\geq\tfrac12\min(p,|S|^2).}
$$
Starting from $S_0=B$, define $S_{j+1}=3S_jS_j-3S_jS_j$. Choose
$$
k=1+\left\lceil\log_2(1/\beta)\right\rceil.
$$
For $p\geq\max(4,2^{2/\beta})$, the growth lemma implies $|S_k|\geq p/2$. To see the constants explicitly, put $v_j=|S_j|/2$. Then $v_0\geq p^{\beta/2}$ and $v_{j+1}\geq\min(p/4,v_j^2)$. Induction gives $v_j\geq\min(p/4,p^{\beta2^{j-1}})$, and the chosen $k$ makes the exponent at least one.
Every $t\in S_0$ has $D(t)\leq K$. If $D(t)\leq K^{L_j}$ on $S_j$, the closure bounds show that each product has exponent at most $3L_j+5$, its negative at most $3L_j+7$, and a sum of six such terms has exponent at most $18L_j+67$. Thus, with
$$
L_0=1,\qquad L_{j+1}=18L_j+67,
$$
we have $|A+tA|\leq K^{L_k}n$ for every $t\in S_k$. The <integer> $L_k$ depends only on $\beta$.
Now apply the collision count to $A$ instead of $S$. Summing over all <scalars> gives at most $pn^2+n^4$. Since $|S_k|\geq p/2$, some $t\in S_k$ has collision count at most $2n^2+2n^4/p$. The <Cauchy-Schwarz inequality> therefore gives
$$
K^{L_k}n\geq|A+tA|\geq\frac{n^2}{2+2n^2/p}\geq\frac14\min(n^2,p).
$$
It follows that
$$
K^{L_k}\geq\frac14\min(n,p/n)\geq\frac14p^\alpha.
$$
For $p\geq4^{2/\alpha}$ this gives $K\geq p^{\alpha/(2L_k)}\geq n^{\alpha/(2L_k)}$, proving the desired positive exponent for all sufficiently large primes.
For completeness, the same conclusion with a possibly smaller exponent holds for the remaining primes. Choose an <integer>
$$
P_0\geq\max(4,2^{2/\beta},4^{2/\alpha}),\qquad
c_{\alpha,\beta}=\min\left(\frac{\alpha}{2L_k},\frac{\log(1+1/P_0)}{\log P_0}\right)>0.
$$
If $p<P_0$, choose any nonzero $b\in B$. We have $1<n<p$. If $|A+bA|=n$, all its translates $A+u$ for $u\in bA$ would coincide. Two distinct such $u$ would make $A$ invariant under a nonzero <translation>, forcing $A=\mathbb F_p$, a contradiction. Thus $|A+bA|\geq n+1\geq n^{1+c_{\alpha,\beta}}$ by the definition of the second exponent. Together with the large-prime case this proves
$$
\boxed{\exists b\in B:\ |A+bA|\geq|A|^{1+c_{\alpha,\beta}}.}
$$
Positivity of the parameters cannot be dropped: at $\alpha=0$, taking $A=\mathbb F_p$ prevents any expansion; at $\beta=0$, the allowed set $B=\{0\}$ prevents expansion of any proper $A$ with more than one element. No external sum-product theorem was invoked: the <scalar> growth and transfer arguments were proved above.
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