If $n\ge 2m+2$ then the generator of $H^{2m+2}({\mathbb C}P^n/{\mathbb C}P^m)$ has a non-trivial cup square, which easily implies that the space is not co-reducible. This leaves open the cases when $(m,n)$ is one of the following pairs $(1,3), (3,5), (3,6)$, or $(3,7)$. Let us consider them one by one.
Consider the action of mod $3$ Steenrod algebra on $H^*({\mathbb C}P^\infty; {\mathbb Z}/3)\cong {\mathbb F}_3[x]$. An easy calculation shows that ${\mathcal P}^1(x^4)=x^6$. It follows that in ${\mathbb C}P^{6}/{\mathbb C}P^3$ and ${\mathbb C}P^7/{\mathbb C}P^3$ the bottom cell (in dimension 8) is connected to the cell in dimension 12 by a Steenrod operation, and so these spaces are not co-reducible.
${\mathbb C}P^3/{\mathbb C}P^1$ and ${\mathbb C}P^5/{\mathbb C}P^3$ are two-cell complexes determined by attaching maps $S^5\to S^4$ and $S^9\to S^8$ respectively. Both sets of homotopy classes $[S^5, S^4]$ and $[S^9, S^8]$ have just one non-trivial element, which is the suspension of the Hopf map $S^3\to S^2$. The cofiber of the Hopf map has a non-trivial action of $Sq^2$. But $Sq^2$ acts trivially on $x^2$ and $x^4$ in the cohomology of ${\mathbb C}P^\infty$, which means that the attaching maps in our cases are nullhomotopic, and so these two spaces are in fact co-reducible.
Finally for the quaternionic space you also can use the action of ${\mathcal P}^1$ in the mod $3$ Steenrod algebra to show that the cell in dimension 8 is connected to the cell in dimension 12, and so ${\mathbb H}P^n/{\mathbb H}P^1$ is never co-reducible.