Dimensional property of kernel for sum of two linear maps

Let $$T: V \longrightarrow V, S: V \longrightarrow V$$ be two linear operators. Let $$P: V \longrightarrow V$$ be another linear operator. Suppose $$P \circ S=S \circ P$$, then prove or disprove that $$\operatorname{dim}(\operatorname{ker}(T \circ S+P))=\operatorname{dim}(\operatorname{ker}(S \circ T+P))$$

Is there any isomorphism possible from the space $$R_1=\{T(S(x))+P(x):~x \in V\}$$ to $$R_2=\{S(T(y))+P(y): y \in V\}$$ using the fact $$P(S(x))=S(P(x))$$ so that we can apply Rank-Nullity theorem? Is there any result regarding $$\dim \ker (T+P)$$ for linear maps $$T$$ and $$P$$?

This is not true. Counterexample: $$T=\left(\begin{array}{cc|cc}0&0&1&0\\ 0&0&0&1\\ \hline 1&0&0&0\\ 0&1&0&0\end{array}\right), \ S=\left(\begin{array}{cc|cc}0&0&0&0\\ 0&1&0&0\\ \hline 0&0&0&1\\ 0&0&0&0\end{array}\right), \ P=\left(\begin{array}{cc|cc}0&0&0&0\\ 0&0&0&0\\ \hline 0&0&1&0\\ 0&0&0&1\end{array}\right),$$ $$TS+P=\pmatrix{0&0&0&1\\ 0&0&0&0\\ 0&0&1&0\\ 0&1&0&1}, \ ST+P=\pmatrix{0&0&0&0\\ 0&0&0&1\\ 0&1&1&0\\ 0&0&0&1}.$$ We have $$PS=SP$$ but $$\operatorname{rank}(TS+P)=3\ne2=\operatorname{rank}(ST+P)$$.
Note, however, if $$S$$ is assumed to be invertible, then the statement is true:
If $$S$$ is invertible, then $$T\circ S+P$$ has same rank as $$(T\circ S+P)\circ S^{-1}=T+P\circ S^{-1}=T+S^{-1}\circ P$$ which has the same rank as $$S\circ (T+S^{-1}\circ P) =S\circ T+P$$.