So from the general reaction we can see that the carbonyl is reduced while the Grignard is oxidized to an alkene.
So for this hypothetical reaction (does this really occur?) we are asked to provide a plausible mechanism. This is what I got, but the prof. wants something better - why transfer negative charge from oxygen to an $\text{sp}^3$ carbon?
Click on the image to enlarge, if you wish.
Here are his hints:
Remember, prior to Grignard reaction the $\ce{Mg}$ center swaps out an ether solvent molecule for a reactant oxygen. This interaction will lead to the reaction but in this case, due to steric interference a hydride is delivered instead as the Grignard is oxidized to an alkene. In this way the Grignard reagent is almost reacting in the way $\ce{NaBH4}$ would in the presence of this ketone. Try to get all of this action working together in a single, concerted step.
So ... I get the first part just fine. The $\ce{Mg}$ center has an octet; it's solvated by two ether molecules in solution usually. Okay and in this case the carbonyl oxygen helps solvate the $\ce{Mg}$ center. And then hydride transfer occurs ... but how exactly do we get the hydride to leave without creating a carboanion? And is it really that important that everything be concerted?
This is my second attempt ... if this is all concerted in real life then I must have hit the orbital alignment lottery! All the stars must have been aligned ... !
The only thing I see about my second attempt is that in the end the $\ce{Mg}$ center doesn't have a complete octet - perhaps I should have shown another ether molecule popping in to help $\ce{Mg}$ out with its octet fantasies.