The hypotheses you have now don't seem to rule out that here are 2 states $i\ne j$ with $P_{i,j}=1$ and $P_{j,i}=1$. In this case $P^n$ does not converge because $P^n_{i,i}$ is 0 or 1 depending on whether $n$ is odd or even.

Suppose the answer is Yes. Then suppose we add two more states $i\ne j$ with $P_{i,j}=1$ and $P_{j,i}=1$, and no other state can go to states $i$ or $j$. Then for the new matrix the assumptions are still satisfied, but now the answer is No. Therefore the answer must be No.

Yes, $P^n$ converges to the matrix Q with $Q_{i,j}=0$ if $j\ne\text{null}$ and $Q_{i,j}=1$ if $j=\text{null}$. This is because $P^n_{i,j}$ is the probability that if started in $i$ after $n$ moves we are in state $j$. Take $\lambda>0$ such that $P_{i,\text{null}}\ge\lambda$ for all $i$. Then $P^n_{i,j}\le (1-\lambda)^n\longrightarrow 0$ for all $j\ne\text{null}$.

(Note that the question is not about almost sure behavior but about a limit of matrices, so Borel-Cantelli is not relevant to the question as stated.)

The hypotheses you have now don't seem to rule out that here are 2 states $i\ne j$ with $P_{i,j}=1$ and $P_{j,i}=1$. In this case $P^n$ does not converge because $P^n_{i,i}$ is 0 or 1 depending on whether $n$ is odd or even.