Let $F_K(\overline{X})$ be the $K$-free algebra over $\overline{X}$. I want to prove that $F_K(\overline{X})\in ISP(K)$.
I have already proved that $F_K(\overline{X})\in IP_SIS(K)$. Since $P_S\leq SP$, we have $F_K(\overline{X})\in ISPIS(K)$. How can I conclude that $F_K(\overline{X})$ actually lies in $ISP(K)$?
The only fact I know about $K$ is that $K\neq\varnothing$.
I claim $ISPIS(K) = ISPS(K) = ISP(K)$.
First equality: Clearly $ISPS(K) \subseteq ISPIS(K)$. Conversely, if $A$ is in $ISPIS(K)$, it is isomorphic to a substructure of a product of some algebras $(B_i)_{i\in I}$, such that each $B_i$ is isomorphic to a structure $C_i$, which is a substructure of something in $K$. But then $A$ is also isomorphic to a substructure of the product of the $C_i$ (its image under the isomorphism $\prod_I B_i \cong \prod_I C_i$), so $A$ is in $ISPS(K)$.
More generally, as long as you close under isomorphism at the end, it's irrelevant whether you've closed under isomorphism somewhere in the middle.
Second equality: It suffices to show $SPS(K) = SP(K)$, and clearly $SP(K)\subseteq SPS(K)$. Suppose $A\in SPS(K)$, i.e. $A$ is a substructure of $\prod_I B_i$, where each $B_i$ is a substructure of some $C_i\in K$. But then $\prod_I B_I$ is a substructure of $\prod_I C_i$, so $A$ is a substructure of $\prod_I C_i$, and $A\in SP(K)$.