There is no solution.
Proof by induction
Here is one-line summary: the bottom string is always longer than the top string by a string in the form $(abca\mid abcaaa\mid abcaaab)^+$.
Suppose we try to construct a solution by adding dominos one by one, keeping the string on the top row (the top string) and the strong on the bottom row (the bottom string) consistent all the time. "Consistent" means either the top string is a prefix of the bottom string or the bottom string is a prefix of the top string.
Initially both strings are empty. Then, however far we may try, as you have noticed, the bottom string is always longer than the top string.
However, we cannot prove that simple statement by straightforward induction since, when considered as the induction hypothesis, it is too weak to support the induction step.
Let us observe the overhang, the extra part of the bottom string. It changes as follows:
$b\to ca \to a\to ab\to bab\to abca\to abcaa\to\cdots$
Continuing, we find the overhang grows as follow:
$$\begin{aligned}
&\to^*\underline{abca}\\
&\to^*\underline{abcaa}\\
&\to^*\underline{abcaaab}\\
&\to^*\underline{abcaaab}\,\underline{abca}\\
&\to^*\underline{abca}\,\underline{abcaaab}\,\underline{abca}\\
&\to^*\underline{abcaaab}\,\underline{abca} \,\underline{abcaa}\\
&\to^*\cdots
\end{aligned}$$
where $o_1\to^* o_2$ means overhang $o_1$ is transformed to overhang $o_2$ in several steps without ever being empty.
Claim (decomposition of the overhang): The overhang is $abca$ when there are 6 dominos. Suppose the overhang is in the form $(abca\mid abcaa\mid abcaaab)^+$. Then it will be transformed to the same form when $3$, $4$ or $6$ dominos are added.
Proof: In the following $W$ stands for some string.
The overhang is transformed by the following rules.
- If it is $aW$, the next domino can be and can only be
$\begin{bmatrix}a\\ ab\end{bmatrix}$.
The new overhang will be $Wab$.
- If it is $caW$, the next domino can be and can only be
$\begin{bmatrix}ca\\ a\end{bmatrix}$.
The new overhang will be $Wa$.
- If it is $bW$, where $W$ does not start with $ac$, the next domino can be and can only be
$\begin{bmatrix}b\\ ca\end{bmatrix}$.
The new overhang will be $Wab$.
Hence, we can verify that
- if the overhang is $\underline{abca}W$, then it will become $W\underline{abcaa}$ when $3$ dominos are added.
- if the overhang is $\underline{abcaa}W$, then it will become $W\underline{abcaaab}$ when $4$ dominos are added.
- if the overhang is $\underline{abcaaab}W$, then it will become $W\underline{abcaaab}\,\underline{abca}$ when $6$ dominos are added.
With the claim, it is straightforward to use induction to show that the overhang is never empty. That means the bottom string is always longer than the top string. There is no solution to the PCP problem.
Proof by counting
Let us call $\begin{bmatrix}b\\ ca\end{bmatrix},$
$\begin{bmatrix}a\\ ab\end{bmatrix},$
$\begin{bmatrix}ca\\ a\end{bmatrix},$
$\begin{bmatrix}bac\\ c\end{bmatrix}$ type 1, 2, 3, 4 respectively.
Lemma, more type-3 than type-4: Suppose the list of $n$ dominos $d_1, d_2, \cdots, d_n$, $n\ge3$ is a solution. Then there are more type-3 dominos than type-4 dominos in the list.
Proof: The first 3 dominos in the list must be
$$\begin{bmatrix}a\\ ab\end{bmatrix}
\begin{bmatrix}b\\ ca\end{bmatrix}
\begin{bmatrix}ca\\ a\end{bmatrix},$$
where the third domino is $\begin{bmatrix}ca\\ a\end{bmatrix}$ but there is no $\begin{bmatrix}bac\\ c\end{bmatrix}$.
Suppose $d_i$ is type-4. It presents a substring $bac$ on the top row. Since the given list is a solution, it corresponds to a substring $bac$ on the bottom row.
Consider that substring $bac$ on the bottom row. The letter $a$ in the middle is neither preceded by $c$ nor followed by $b$. Hence it can only be brought by some $d_{x(i)}$ of type-3, $1\le x(i)\le n$. $d_i\mapsto d_{x(i)}$ is a correspondence from type-4 dominos to type-3 dominos in the list. Note that a different $d_i$ must be mapped to a different $d_{x(i)}$. Moreover, the third domino, which is type-3 is not mapped to by any type-4 domino.
Claim: There is no solution to this PCP problem.
Proof. Suppose there is a solution. Let $\#1, \#2, \#3, \#4$ be the number of type-1,2,3,4 dominos in the solution. The number of $a$s on the top row and on the bottom row are $\#2+\#3+\#4$ and $\#1 +\#2+\#3$ respectively. The number of $c$s on the top row and the bottom row are $\#3+\#4$ and $\#1+\#4$ respectively. So we have
$$\begin{aligned}
\#2+\#3+\#4 &= \#1+\#2+\#3,\\
\#3+\#4 &= \#1+\#4,
\end{aligned}$$
which implies $\#3=\#4$.
On the other hand, there must be at least $3$ dominos in the solution. The lemma says $\#3>\#4$. This contradiction means we do not have a solution.