The answer is No.
We define two sets of elements of $\{0,1,2\}^\omega$ in the following way:
- For $n\in\omega$ let $u_n$ be defined by $u_n(k)=1$ for $k\leq n$
and $u_n(k)=0$ for $k>n$;
- For $n\in\omega$ let $t_n$ be defined by $t_n(k)=1$ for $k\leq n$
and $t_n(n+1) = 2$ and $t_n(k)=0$ for $k>n+1$;
Note that informally speaking, the $u_n$ have the form $(1,1,1,\ldots, 1,0,0,\ldots)$
and the $t_n$ have the form $(1,1,1,\ldots, 1,2,0,\ldots)$.
Then we set $$E:=\{u_n:n\in \omega\}\cup\{t_n: n\in \omega\}.$$ The ordering
on $E$ is componentwise (equivalently, the ordering inherited from the product
ordering on $\{0,1,2\}^\omega$).
Step 1. Let $c_n = \downarrow t_n$. Then $P = \{c_n: n\in \omega\}$ is a club.
Proof. We have to show that for $m<n $ the tods $c_m, c_n$ are
incompatible. This is the case if we find incomparable elements
in $c_m, c_n$. This is easy: $t_m\in c_m$ and $t_n\in c_n$ are incomparable
in the ordering we chose for $E$.
Step 2. The set $m = \{u_n: n\in \omega\}$ is a maximal tods.
Proof. It's easy to see that $m$ is a chain and a down-set. Next, maximality: if we consider $m\cup\{t_n\}$ for some $n\in \omega$, the elements
$u_{n+2}$ and $t_n$ are not comparable, so $m\cup\{t_n\}$
is not totally ordered.
Step 3. If $x$ is a finite tods with $x\subseteq m$, then
$x$ is compatible with a member of $P$.
Proof. Any finite tods $x\subseteq m$ has the form
$x = \downarrow u_n$ for some $n\in \omega$. So $c_n = \downarrow
t_n \supseteq x$ therefore $c_n \cup x = c_n$ is a tods,
so by definition $c_n$ and $x$ are compatible.
Conclusion. There is no complete club consisting of finite sets $P'\supseteq P$
such that $P$ contains a finite subset of $m$. Therefore the answer to the question is No.
