I haven't found any reference. Moreover, the fact that Mizokami's article

"The embedding of a mapping space with compact open topology", Topol. Appl. 1998

proves it for the pointwise convergence topology on the space $\mathrm{map}(PX,PY)$ together with the fact that no paper citing his paper proves my original question makes me think that it is not written in the literature. Here is a proof:

**The map $\phi\colon \mathrm{map}(X,Y) \to \mathrm{map}(PX,PY)$, where both spaces are endowed with the compact-open topology, is continuous if $X$ is regular.**

*Proof*: It suffices to prove that for each $f\in \mathrm{map}(X,Y)$ and for each subbasic neighbourhood $Q$ of $\phi(f)$, there is a subbasic neighbourhood $P$ of $f$ such that $\phi(P) \subset Q $.

Let us recall first how the topology in the righthandside is defined:

A basic open subset in $PY$ is given by a finite sequence $\langle U_i\rangle$ of open subsets $U_i$ of $Y$. A compact subspace $K\subset PY$ belongs to this open subset if $K\subset \bigcup U_i$ and $K\cap U_i\neq \emptyset$ for all $i$.

By Theorem 2.5 in "Topologies of spaces of subsets", by Ernest Michael, a compact subset ${\mathcal K}$ of $PX$ satisfies that the union $\bigcup_{K\in {\cal K}} K$ is compact as well.

Therefore a basic open subset of $\mathrm{map}(PX,PY)$ is given by pairs $W({\cal K}:\langle U_i\rangle)$ with
$$
W({\cal K}:\langle U_i\rangle) = \{F:PX\to PY\mid F(K)\subset \langle U_i\rangle\} = \{F\mid F(\bigcup_{K\in {\cal K}} K)\subset \bigcup U_i, F(K)\cap U_i\neq \emptyset, \forall K\in {\cal K}, \forall i\}
$$
Let us take such subbasic neighbourhood of $\phi(f)$.

We assert that for each $U'_i:= f^{-1}(U_i)$ there is an open subset $V:=V_i$ such
that $\bar{V}_i\subset U_i$ and for each $K\in {\cal K}$ it holds
that $K\cap V_i\neq \emptyset$.

If this were not true, for each such $V$ there would exist a
$K_{V}\in {\cal K}$ such that $K_V\cap V=\emptyset$. This defines a
net in ${\cal K}$ which has a convergent cofinal subnet because $\cal
K$ is compact. The limit point $K_0\in {\cal K}$ of this subnet
satisfies that $K_0\cap V = \emptyset$ for all such $V$, and since the subnet is
cofinal and $X$ is regular, it also holds that $K_0\cap U_i'=\emptyset$, but this
contradicts the fact that $f(K_0)\cap U_i\neq \emptyset$.

Let us choose $V_i\subset U_i'$ with the above property. Take now a point $p_K^i\in V_i\cap K$ for each $i$ and $K$, and define $K_i$ to be the closure of $\{p_K^i\}$, which is compact because it is a closed subset of $\bigcup_{K\in {\cal K}} K$ (which is itself comapact). By construction it holds that $f(K_i)\subset \bar{V}_i\subset U_i$. Therefore the neighbourhood
$$
W(\bigcup_{K\in {\cal K}} K: \bigcup_i U_i)\cap \bigcap_i W(K_i:U_i)
$$
of $f$ is mapped into the neighbourhood $W({\cal K}:\langle U_i\rangle)$, and therefore $\phi$ is continuous.