$\underline{\textbf{Answer:}\Rightarrow}\;1)\;\text{Only pair (A, B)}$
$\underline{\textbf{Explanation:}\Rightarrow}$
Let $\mathbf A$ be an alphabet with at least two symbols. The input of the problem consists of two finite lists $\mathrm{\alpha_1,\dots,\alpha_N}$ and $\mathrm{\beta_1,\dots,\beta_N}$ of words over $\mathbf A$.
A solution to this problem is a $\mathbf{\color {magenta} {sequence}}$ of indices $\mathrm{(i_k)_{1\le k\le k}}$ with $\mathrm{K \ge 1}$ and $\mathrm{1\le i_k \le N}$ for all $\mathrm k$, such that
$\alpha_{i_1}\dots \alpha_{i_k} = \beta_{i_1}\dots \beta_{i_k}$
The decision problem then is to decide whether such a solution exist or not.
Source:
https://en.wikipedia.org/wiki/Post_correspondence_problem
$\mathbf{ \color {green}{\Large A}}$
A1
A2
A3
A4
001
0011
$\color{magenta}{11}$
$\color{magenta}{101}$
$\mathbf{ \color {green}{\Large B}}$
B1
B2
B3
B4
$\color{magenta}{\enclose{circle}{01}}$
$\color{magenta}{111}$
$\color{blue}{\enclose{circle}{111}}$
010
$\mathbf{ \color {green}{\Large C}}$
C1
C2
C3
00
001
1000
$\mathbf{ \color {green}{\Large D}}$
D1
D2
D3
0
11
011
$\therefore \mathbf{A,\;B}$ are in $\textbf{PCP}$ because:
$\mathbf{\underset{\enclose{circle}{A_3=11,A_4 = 101}}{A_3A_4} = \underset{\enclose{circle}{B_3=111,B_1=01}}{B_3B_1} = 11101}$, and $\mathbf{\underset{\enclose{circle}{A_3 = 11, A_4 = 101}}{A_3A_4} = \underset{\enclose{circle}{B_2=111,B_1=01}}{B_2B_1}=11101}$
$\therefore $ Pair $\mathbf{A,\;B}$ have a solution.
$\therefore (1)$ is the right answer.