At $d)$, minimizer is comparing the $2$ he found with the currently best option for the maximizer higher in the tree (at node $A$, where maximizer will choose between $B$,$C$,...), which is $3$, at node $B$. He then knows that maximizer will not choose $C$.
The node $C$ will have value $2$ when minimizer sets it and would have value $2$ or less if minimizer explored it fully, so he doesn't need to waste his time searching other branches from $C$ if he already knows that maximizer will not choose anything lower than $3$.
I think what confuses you may be the following:
You are imagining a real-life gameplay scenario, but this is different in the sense that minimizer and maximizer must play their best move and will calculate it correctly.
So there is no way that minimizer could in this situation "fool" maximizer by setting the wrong value of $C$ (something higher than $3$) to force maximizer to choose it. If we assume that scenario a possibility, we must also implement the maximizer with "free-will" so he may not choose his best option but this is no longer an $\alpha$-$\beta$ pruning :)