If I connect a cut-off BJT and a diode in series, which circuit element will see a voltage drop?
The simplest models of a diode and transistor can't answer this question. You need to use a more detailed model of your devices to understand what will happen.

In particular, the transistor will allow a small leakage current to flow, even in cut-off mode. From the Fairchiled 2N3904 datasheet:



Of course your circuit can't apply more than 5 V to the collector, so the leakage will be somewhat lower, maybe 10 nA.

10 nA or even 50 nA through the diode will produce almost no voltage drop. 10-50 nA through the 100 ohm resistor will similarly produce only a few microvolts of drop. Thus the bulk of the voltage drop (~5 V) will appear across the transistor's collector and emitter.