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Let me answer as I've learned.

In this case, if the voltage source for forward bias is greater than the voltage source for reverse bias, current will flow in this semiconductive diode.

And another way may occur. That is, in case reverse bias voltage is as large as breakdown voltage, reverse breakdown current ,which is because of the minority carrier in p region, will flow and this current can be large to damage the diode.

If there is any mistake in my answer, please correct me and I'll thank you for that.

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Q: Explain how a current flows in a semiconducting diode in the two states of biasing?
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