We say that a first-order formula is in DNF if it can be obtained from a propositional formula in DNF by replacement of propositional variables with atomic predicate formulas. We say that $A$ is in PNF if $A=\textbf{Q}_1x_1\dots\textbf{Q}_nx_nA'$ where $\textbf{Q}_i$ are quantifiers and $A'$ is in DNF.
Show that for any $A$ there exists a formula $B$ in PNF such that $A\equiv B$ and $B=\exists x_0\forall x_1\dots\exists x_{n-1}\forall x_nB'$ where $B'$ is in DNF.
Since we know that there exits $C=\textbf{Q}_1x_1\dots\textbf{Q}_nx_nC'$ in PNF such that $C\equiv A$ we can just put 'sham' quantifiers between any of $\textbf{Q}_ix_i$ and $\textbf{Q}_{i+1}x_{i+1}$ if $\textbf{Q}_i=\textbf{Q}_{i+1}$ (for instance, if $\textbf{Q}_i=\textbf{Q}_{i+1}=\exists$ we put $\forall y$ between them where $y$ does not occure in $C$). This way we get what we need but the result seems a bit hollow.
If we interpret the statement as "let $C=\textbf{Q}_1x_1\dots\textbf{Q}_nx_nC'$ be a PNF of $A$. There exists a formula $B$ in PNF such that $A\equiv B$ and $B=\exists x_0\forall x_1\dots\exists x_{n-1}\forall x_nB'$ where $B'$ is in DNF" then it looks implausible. For example, consder $A=\forall x\forall yR(x,y)$. This formula is in PNF. I see no way to build it's PNF looks like $\exists x\forall yA',\:A'$ is in DNF.
Maybe I miss something?