$p$-completeness of the function spectrum $F(\Sigma^{\infty} BS, \Sigma^{\infty} BK)$

Tim's argument is correct, and here's a different way to see this.

To say that $\Sigma^{\infty} BS$ is a finite $p$-group has trivial rationalization and is $p$-local is the same as to observe that it is $p$-torsion, i.e. has $p$-torsion homotopy groups.

To see this without involving the Serre spectral sequence, observe that both the rational and $\mathbb{Z}/l$ cohomology (for $l \neq p$) of $\Sigma^{\infty} BS$ vanishes by Maschke's theorem. For a connective spectrum, this forces it to be $p$-torsion, by Hurewicz (applied to $\Sigma^{\infty} BS \otimes M(l)$).

We also have that $F(T, U)$ is p-complete for any $p$-torsion $T$ and arbitrary $U$.

To see the latter, observe that the subcategory of $p$-torsion spectra is generated under colimits and desuspensions by $M(p)$ (this is the same as saying that any such non-zero spectrum admits a non-zero map from $M(p)$, which is clear). Since $F(-, U)$ takes colimits to limits, and $p$-complete spectra are closed under limits, we then just need to know $F(S^{0}/p, U)$ is $p$-complete, but this is the same as $\Sigma^{-1} U \otimes S^{0}/p$, so we're done.

Here is an argument (not as clean as Piotr's below). Use with caution; it's possible that I've made a mistake. We don't use any properties of $\Sigma^\infty BK$ -- this could be an arbitrary spectrum. But it's crucial that that we use $\Sigma^\infty BS$ with $S$ a finite $p$-group.

Proposition: If $S$ is a finite $p$-group and $U$ is any spectrum, then $F(\Sigma^\infty BS, U)$ is $p$-complete.

We will prove this using the following lemmas:

Lemma 1: If $S$ is a finite $p$-group, then $\tilde H_\ast(BS; A) = 0$ if $p: A \to A$ is an isomorphism.

As Piotr points out in his answer, this is a consequence of Maschke's theorem. We provide a proof using basic algebraic topology and finite group theory.

Proof: We reduce to the case where $S$ is abelian, where this is a standard calculation. For if $S$ is nonabelian, then there is always a nontrivial short exact sequence $Z(S) \to S \to S / Z(S)$ where $Z(S)$ is the center of $S$. Because $Z(S)$ is the center of $S$, the action of $S/Z(S)$ on $Z(S)$ is trivial. Thus in the Serre spectral sequence $H_\ast(B(S/Z(S)); \underline{H_\ast(B(Z(S))}) \Rightarrow H_\ast(BS)$ we have trivial coefficients. So we can induct on the order of $S$.

Corollary 2: Let $S$ be a finite $p$-group. Then $\Sigma^\infty BS$ is $p$-local.

Proof: Let $\ell \neq p$ be a different prime. The claim is that $\ell: \Sigma^\infty BS \to \Sigma^\infty BS$ is an equivalence of spectra. It suffices to show that $\ell: \Sigma BS \to \Sigma BS$ is an equivalence of spaces. By the homology Whitehead theorem for field coefficients, it suffices to show that $\ell: \tilde H_\ast(BS;k) \to \tilde H_\ast(BS;k)$ is an isomorphism for $k = \mathbb Q$ or $k = \mathbb F_q$ with $q$ a prime. But if $k = \mathbb Q$ or $k = \mathbb F_q$ with $q \neq p$, both sides are zero by Lemma 1, while if $k = \mathbb F_p$, then this follows from $\ell$ being prime to $p$.

Lemma 3: Let $T$ be a $p$-torsion spectrum -- i.e. $T$ is $p$-local and has trivial rationalization -- and let $U$ be an arbitrary spectrum. Then $F(T,U)$ is $p$-complete.

Proof: Let $X$ be such that $X \wedge M(p) = 0$; the claim is that $Map(X,F(T,U)) = 0$. Equivalently, the claim is that $Map(T, F(X,U)) = 0$. Because $T$ is $p$-local, it is equivalent to claim that $Map(T, F(X,U)^{(p)}) = 0$, where we have taken a $p$-colocalization (i.e. we have applied the right adjoint $(-)^{(p)}$ to the inclusion of the $p$-local spectra into all spectra). Then, $$F(X,U)^{(p)} \wedge M(p) = (F(X,U)\wedge M(p))^{(p)} = F(X \wedge \Sigma^{-1} M(p), U)^{(p)} = 0$$ The first equivalence comes because $(-)^{(p)}$ commutes with finite colimits, the second by Spanier-Whitehead duality, and the third by the hypothesis that $X \wedge M(p) = 0$. Since $F(X,U)^{(p)}$ is by definition $p$-local, this means that it is rational. So $$Map(T, F(X,U)^{(p)}) = Map_{H\mathbb Q}(H\mathbb Q \wedge T, F(X,U)^{(p)}) = 0$$ because by hypothesis $H\mathbb Q \wedge T = 0$.

Proof of Proposition: By Lemma 1, $\Sigma^\infty BS$ has trivial rationalization, and by Corollary 2, $\Sigma^\infty BS$ is $p$-local. So by Lemma 3, $F(\Sigma^\infty BS, U)$ is $p$-complete.