Solution:

Answer: $20\pi d\sqrt{(\alpha\beta)}$ kilometers per hour.

Let the plane be at height $h$ and a (horizontal) distance $y$ from $A$. Let the angle $PAB$ be $\theta + \alpha$ and the angle $PBA$ be $\phi$. After 1 second, the angle $PAB$ is $\theta$ and the angle $PBA$ is $\phi + \beta$. We have immediately that:

$$h / y = \tan(\theta + \alpha),\quad h / (d - y) = \tan\phi,\quad h / (y + x) = \tan\theta,\quad h / (d - y - x) = \tan(\phi + \beta).$$

Eliminating $\theta$, we obtain:
$$h / y = (\tan\alpha + \tan\theta) / (1 - \tan\alpha \tan\theta) = (a(y + x) + h) / (y + x - ah)$$
where $a = \tan\alpha$. Hence
$$x = a(h^2 + y^2) / (h - ay).$$
Similarly, eliminating $\phi$, we obtain
$$x = b(h^2 + (d - y)^2) / (h + (d - y)b).$$

At this point I do not see how to make further progress without approximating. But approximating seems reasonable, since $\alpha$ and $\beta$, are certainly small, at least when expressed in radians. For example, typical values might be 10,000 ft for $h$ and more than 10 miles for $y$ or $d- y$ and 500 mph for the aircraft speed. That gives $x = 0.14$ miles, so $x / y = 0.014$ and $x / h = 0.07$. So, let us neglect $a / h$, $b / h$, $a / y$ etc. Then we get the simplified expressions:
$$x = a(h^2 + y^2) / h = b(h^2 + (d - y)^2) / h.$$

If $a = b$, then we quickly obtain $y = d / 2$, $h = d / 2$, $x = ad$. Assume $a > b$. Then we can solve for $h$, substitute back in and obtain an expression for $x$ in terms of $y$. It is convenient to divide through by $d$ and to write $X = x / d$, $Y = y / d$. Note that since we are assuming $a > b$, we require $Y < 1 / (1 + \sqrt{(a / b)})$. After some manipulation we obtain:
$$X = ab(1 - 2Y) / \sqrt{((a - b)(b(1 - Y)^2 - aY^2))}.$$
Differentiating, we find that there is a minimum at $Y = b / (a + b)$, which is in the allowed range, and that the minimum value of $X$ is $\sqrt{(ab)}$. By symmetry, we obtain the same result for $a < b$ and we notice that it is also true for $a = b$. So in all cases we have that the minimum value of $x$ is $d\sqrt{(ab)}$.

We are assuming $\alpha$ and $\beta$ are small, so we may take $a = \alpha$, $b = \beta$. However, the question specified that $\alpha$ and $\beta$ were measured in degrees, so to obtain the final answer we must convert, giving:
$$x = d(\pi / 180) \sqrt{(\alpha\beta)},$$
and hence
$$speed = 20\pi d\sqrt{(\alpha\beta)}$$
kilometers per hour.