First, notice that the two greeks you mentioned in your question are simply the partial derivatives of the value of the option $V$ with respect to two different variables $S$ (the price of the underlying) and $\sigma$ (the volatility of the underlying):
$$\Delta = \frac{\partial V}{\partial S} \quad \text{and} \quad \nu=\frac{\partial V}{\partial \sigma}$$
As mentioned in the comments, volatility is not per-say a sign of declining prices, but, at least under the Black-Scholes model, $\nu$ is positive for both puts and calls meaning that the value of both types of options is expected to rise if $\sigma$ goes up.
So, from a mathematical point of view, if only $\sigma$ goes up (i.e. $S$ stays the same) then yes the trade would be profitable even by buying a put option. However, it is fair to say that this situation is not very realistic and that in real-world you would be exposed to wild moves in $S$ which would affect the price of the option through the $\Delta$.
As a result, what you would like to do is to is to perform delta hedging which consists in offsetting your exposure to $S$ by buying or selling an amount of the underlying $S$ corresponding to $\Delta$. In the case of the put, you know that $\Delta<0$ (i.e. the price of the put $V$ decreases as the price of the underlying $S$ increases) and you hence need to buy $\Delta$ of $S$ to make your resulting portfolio have a $\Delta=0$: it is delta-neutral.
Once you've done that, you've removed your exposure to changes in $S$ and you could consider that your put position is mainly determine by the remaining $\nu$. Some funds such as the Amundi Volatility Funds do exactly that.
This reasoning though is really perfectly working only for small changes in $S$, as bigger changes would also be sensitive to other greeks such as $\Gamma = \frac{\partial^2V}{\partial S^2}$. You can also hedge against this kind of move using a similar reasoning.
