No. Such a map would give rise to a map $PSU(4)/(\mathbb Z/2) \to SU(2)$ inducing multiplication by $d$ on $H^3$ and hence to a map $SU(4) \to SU(2)$ inducing multiplication by $2d$ on $H^3$ and such a map can not exist.

Indeed, let $\Sigma \mathbb{C}P^3 \to SU(4)$ be the axial map (see for instance I.M.James, "The topology of Stiefel manifolds", LMS, p. 22, where the map is called $\phi$) inducing an isomorphism on homology in degree $\leq 7$.
Localizing at $2$ for convenience, if the map $SU(4) \to SU(2)$ as above were to exist then we would be able (2-locally) to extend a degree 2 map $S^3 \to S^3$ to $\Sigma \mathbb{C}P^3$. This is not possible even stably.

$2$-locally the stable homotopy groups of spheres are $\pi_1 S^0 \cong \mathbb{Z}/2 \eta, \ \pi_2 S^0 \cong \mathbb{Z}/2 \eta^2, \ \pi^3 S^0 \cong \mathbb{Z}/8 \nu$ with $\eta^3=4\nu$. Looking at the stable homotopy long exact sequence of the cofiber sequence
$$
S^1 \xrightarrow{\eta} S^0 \to \Sigma^{-2}\mathbb{C}P^2
$$
we see that $\pi_3 \Sigma^{-2}\mathbb{C}P^2 \cong \mathbb{Z}/4$, generated by
$$
S^3 \xrightarrow{\nu} S^0 \to \Sigma^{-2}\mathbb{C}P^2
$$
The attaching map $S^3 \to \Sigma^{-2}\mathbb{C}P^2$ of the top cell in $\Sigma^{-2}\mathbb{C}P^3$ must be $2\nu$: indeed it can't be an odd multiple of $\nu$ because the action of the Steenrod square $Sq^4$ on $H^0(\Sigma^{-2}\mathbb{C}P^3;\mathbb{Z}/2)$ is trivial and it can't be $0$ for then the top cell would split off which is contradicted by the behaviour of Adams operations on complex $K$-theory - see for instance Adams, "Vector fields on spheres", Annals of Math (1962) Theorem 7.2.

Writing $t\colon \Sigma^{-2}\mathbb{C}P^2 \to S^0$ for the map which has degree $2$ on the bottom cell we see that the composite
$$S^3 \xrightarrow{2\nu} \Sigma^{-2}\mathbb{C}P^2 \xrightarrow{t} S^0$$
will therefore equal $0\neq 4\nu \in \pi_{3}S^0$ and therefore the
map $\Sigma^{-2}\mathbb{C}P^2 \xrightarrow{t} S^0$ can not possibly extend to
$\Sigma^{-2}\mathbb{C}P^3$.