Is this equivalent to : all sets are $P\ \textit{definable}$?

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Informally by $P$ definable set it is meant a set that is definable by a formula all parameters of which satisfy a one place predicate symbol $P$, the well known example is "ordinal definable set".

Now is the statement: $$\forall x \in V \ (x \text{ is } P\ \textit{definable} \text{ set})$$ is equivalent to saying that we have a parameter free definable injection from $V$ to the class of all sets satisfying $P$?

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Yes, this is correct.

One direction is trivial. In the other direction, and a bit informally, if $f$ is a parameter-freely definable injection from $V$ to any class $C$ then every element $u$ of $V$ is definable from a parameter $\alpha$ in $C$ - namely, its $f$-image. If $\alpha$ in turn is definable by some $\beta_1,...,\beta_n$, we can "compose definitions" and get a definition of $u$ from the $\beta_i$s.