To make notation a little bit lighter I am going to use $N=(N(t)_1,N(t)_2)$, $X=(X(t)_1,X(t)_2)$, $X'=(X'(t)_1,X'(t)_2)$.

In order to answer the question, I think we need to know what the dependency between $X$ and $N$ is?

Whatever their dependency, the only case when $Pr(X'=x') \neq 0$ is when $N=0$. If not $X'$ is continuous, because $Y$ is continuous, and $Pr(X'=x')=0$.

I hope this helps. Have you tried doing some simulations?

To make notation a little bit lighter I am going to use $N=(N(t)_1,N(t)_2)$, $X=(X(t)_1,X(t)_2)$, $X'=(X'(t)_1,X'(t)_2)$. $x$ is any of the nine possible answers.

The only case when $Pr(X'=x) \neq 0$ is when $N=0$. If not $X'$ is continuous, because $Y$ is continuous, and $Pr(X'=x)=0$.

$ Pr(X'=x) = Pr(X'=x|N=0)Pr(N=0) + Pr(X'=x|N\neq 0)Pr(N\neq 0) = Pr(X'=x|N=0)Pr(N=0) + 0 = Pr(X=x|N=0)Pr(N=0) = Pr(X=x)Pr(N=0)$

Step 1: $Pr(A) = Pr(A \cap B) + Pr(A\cap B^{\complement})$

Step 2: we know that when $N\neq 0$ $Pr(X'=x|N \neq 0)=0$, as explained above.

Step 3: if $N=0$ then $X'=X$

Step 4: $N$ and $X$ are independent. $Pr(N=0)$ can be calculated using the link you provided.