$\sf ZF$ can prove that $\mathcal P(\omega)$ exists, but it cannot prove whether or not every subset of $\omega$ is constructible. In other words, there are sets which provably exist, but it is not provable that they are constructible. 

At the same time, as Mohammad writes, if one can prove that there exists a non-constructible set, then one disproves $V=L$, which means that $\sf ZF$ is already inconsistent to begin with.

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Also, it is not true that $\mathcal P_{\rm def}$ is the "right" power set, since if you look at the $L$-hierarchy, you'll find that new subsets of $\omega$ are added in unboundedly many countable steps below $\omega_1$. So it is not right to say that $\mathcal P_{\rm def}(\omega)$ is "the right power set of $\omega$", if we can prove that the "second power set also adds subsets".