For fixed integers $r,\ell \geq 0$, a graph $G$ is called an {\em $(r,\ell)$-graph} if the vertex set $V(G)$ can be partitioned into $r$ independent sets and $\ell$ cliques. The class of $(r, \ell)$ graphs generalizes $r$-colourable graphs (when $\ell =0)$ and hence not surprisingly, determining whether a given graph is an $(r, \ell)$-graph is \NP-hard even when $r \geq 3$ or $\ell \geq 3$ in general graphs. When $r$ and $\ell$ are part of the input, then the recognition problem is NP-hard even if the input graph is a perfect graph (where the {\sc Chromatic Number} problem is solvable in polynomial time). It is also known to be fixed-parameter tractable (FPT) on perfect graphs when parameterized by $r$ and $\ell$. I.e. there is an $f(r+\ell) \cdot n^{\Oh(1)}$ algorithm on perfect graphs on $n$ vertices where $f$ is some (exponential) function of $r$ and $\ell$. In this paper, we consider the parameterized complexity of the following problem, which we call {\sc Vertex Partization}. Given a perfect graph $G$ and positive integers $r,\ell,k$ decide whether there exists a set $S\subseteq V(G)$ of size at most $k$ such that the deletion of $S$ from $G$ results in an $(r,\ell)$-graph. We obtain the following results: \begin{enumerate} \item {\sc Vertex Partization} on perfect graphs is FPT when parameterized by $k+r+\ell$. \item The problem does not admit any polynomial sized kernel when parameterized by $k+r+\ell$. In other words, in polynomial time, the input graph can not be compressed to an equivalent instance of size polynomial in $k+r+\ell$. In fact, our result holds even when $k=0$. \item When $r,\ell$ are universal constants, then {\sc Vertex Partization} on perfect graphs, parameterized by $k$, has a polynomial sized kernel. \end{enumerate}