Well, in one sense, this is always true. If $Q\subset SP(M)$ is a principal right $H$-bundle, where $H\subset\mathrm{Spin}(n)$ is a subgroup, then $\delta(Q)\subset F(M)$ is a principal right $\pi(H)$-bundle, where $\pi(H)\subset\mathrm{SO}(n)$ is the image subgroup. (Here, I am using $\delta:SP(M)\to F(M)$ to denote the double cover that the OP didn't name or notate.)

The OP may, however, have intended to ask a different question, such as, "When does having a nonvanishing spinor field on $M$ imply a reduction of structure group of $M$?". The answer to this is "When $\mathrm{Spin}(n)$ acts transitively on the unit sphere in its spinor representation $\mathbb{S}$.". (However, one must be precise about that is meant by this. People sometimes say "$\mathrm{Spin}(8)$ acts transitively on the unit sphere in its spin representation." when they mean "$\mathrm{Spin}(8)$ acts transitively on the unit sphere in each of its semi-spin representations.")

**Added at the request of the OP:** Let $(M,g)$ be an oriented, spinnable Riemannian $d$-manifold, with $\beta: F(M)\to M$ being its oriented orthonormal frame bundle and $\delta:SP(M)\to F(M)$ being a double cover that defines a spin structure on $M$. Let $\pi:\mathrm{Spin}(d)\to\mathrm{SO}(d)$ be the double cover and let $\mathbb{S}$ be the spinor space, with $\rho:\mathrm{Spin}(d)\to\mathrm{SO}(\mathbb{S})$ the spin representation. Then $S = SP(M)\times_\rho\mathbb{S}$ is the spinor bundle of the spin structure, so that a spinor field, i.e., a section of $S\to M$ is represented by a smooth map $s:SP(M)\to \mathbb{S}$ that satisfies $s(u\cdot g) = \rho(g^{-1})s(u)$ for all $u\in SP(M)$ and $g\in \mathrm{Spin}(n)$. The section is nonvanishing if and only if $s$ is nonvanishing, in which case, we can normalize, replacing $s$ by $s/|s|$ to get a unit spinor field. Thus, we can assume $|s| = 1$, i.e., that $s$ maps $SP(M)$ to the unit sphere in $\mathbb{S}$. Now, we can ask, when can we find an element $v\in\mathbb{S}$ with $|v|=1$, such that $s^{-1}(v)\subset SP(M)$ defines a principal subbundle of $SP(M)$ for some subgroup $H\subset \mathrm{Spin}(d)$. Well, the only possibility for $H$ would be the $\rho$-stabilizer of $v$, i.e., the subgroup $H = \{ g\in \mathrm{Spin}(d)\ |\ \rho(g)v = v\}$. But this will only work if $s$ takes values in the orbit $\rho\bigl(\mathrm{Spin}(d)\bigr){\cdot}v$. Now, this will always be true if $\mathrm{Spin}(d)$ acts transitively on the unit sphere in $\mathbb{S}$. Unfortunately, that happens only when $d\in\{2,3,5,6,7,9\}$. In the two cases $d = 4$ and $d=8$, you don't have transitivity because, as a representation of $\mathrm{Spin}(d)$, the space $\mathbb{S}$ is the direct sum of two irreducible sub-representations: $\mathbb{S}=\mathbb{S}_+\oplus\mathbb{S}_-$, corresponding to the fact that a spinor field is the sum of two 'semi-spinors'. It turns out that, in these two cases, $\mathrm{Spin}(d)$ does act transitively on the unit spheres in $\mathbb{S}_\pm$, so you get a structure reduction when there is a nonvanishing semi-spinor.