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Let $P=(p_{ij})$ and $Q=(q_{ij})$ be the $n\times n$ transition matrices for the two respective Markov chains. Take $n=5$, $$P=\frac1{1000}\left( \begin{array}{ccccc} 1000 & 0 & 0 & 0 & 0 \\ 1 & 241 & 260 & 38 & 460 \\ 22 & 75 & 283 & 448 & 172 \\ 389 & 67 & 103 & 158 & 283 \\ 0 & 0 & 0 & 0 & 1000 \\ \end{array} \right), $$ $$Q=\frac1{1000}\left( \begin{array}{ccccc} 1000 & 0 & 0 & 0 & 0 \\ 4 & 241 & 259 & 37 & 459 \\ 23 & 531 & 283 & 1 & 162 \\ 390 & 197 & 194 & 158 & 61 \\ 0 & 0 & 0 & 0 & 1000 \\ \end{array} \right). $$

Then $f_{P;21}=\dfrac{30684666}{198426719}=0.154\ldots\not\le0.054\ldots= \dfrac{4510572}{83295877}=f_{Q;21}$, where $f_{P;ij}$ is the probability that the first chain ever reaches a state $j$ from $i$, and $f_{Q;ij}$ is defined similarly.

This disproves your conjecture, because there is no state to the left of $1$.

Remark. Using the formula $f=\lim_{t\uparrow1}(I-tR)^{-1}g$, we see that, for $n=5$, $$f_{P;21}=\big[-p_{24} \left(\left(1-p_{33}\right) p_{41}+p_{31} p_{43}\right)-p_{23} \left(p_{34} p_{41}+p_{31} \left(1-p_{44}\right)\right)+p_{21} \left(-p_{44} p_{33}+p_{33}+p_{34} p_{43}+p_{44}-1\right)\big] \\ \big/\big[p_{23} p_{32}+p_{24} p_{43} p_{32}-p_{23} p_{44} p_{32}+p_{33}+p_{24} p_{42}-p_{24} p_{33} p_{42}+p_{23} p_{34} p_{42}+p_{34} p_{43}-p_{33} p_{44}+p_{44}-p_{22} \left(p_{34} p_{43}-p_{33} \left(p_{44}-1\right)+p_{44}-1\right)-1\big]. $$ This expression depends on the $p_{ij}$'s in a very complicated way, which makes simple comparisons such as the one in the OP unlikely to hold.

Let $P=(p_{ij})$ and $Q=(q_{ij})$ be the $n\times n$ transition matrices for the two respective Markov chains. Take $n=5$, $$P=\frac1{1000}\left( \begin{array}{ccccc} 1000 & 0 & 0 & 0 & 0 \\ 1 & 241 & 260 & 38 & 460 \\ 22 & 75 & 283 & 448 & 172 \\ 389 & 67 & 103 & 158 & 283 \\ 0 & 0 & 0 & 0 & 1000 \\ \end{array} \right), $$ $$Q=\frac1{1000}\left( \begin{array}{ccccc} 1000 & 0 & 0 & 0 & 0 \\ 4 & 241 & 259 & 37 & 459 \\ 23 & 531 & 283 & 1 & 162 \\ 390 & 197 & 194 & 158 & 61 \\ 0 & 0 & 0 & 0 & 1000 \\ \end{array} \right). $$

Then $f_{P;21}=\dfrac{30684666}{198426719}=0.154\ldots\not\le0.054\ldots= \dfrac{4510572}{83295877}=f_{Q;21}$, where $f_{P;ij}$ is the probability that the first chain ever reaches a state $j$ from $i$, and $f_{Q;ij}$ is defined similarly.

This disproves your conjecture, because there is no state to the left of $1$.

Remark. Using the formula $f=\lim_{t\uparrow1}(I-tR)^{-1}g$, we see that, for $n=5$, $$f_{P;21}=\big[-p_{24} \left(\left(1-p_{33}\right) p_{41}+p_{31} p_{43}\right)-p_{23} \left(p_{34} p_{41}+p_{31} \left(1-p_{44}\right)\right)+p_{21} \left(-p_{44} p_{33}+p_{33}+p_{34} p_{43}+p_{44}-1\right)\big] \\ \big/\big[p_{23} p_{32}+p_{24} p_{43} p_{32}-p_{23} p_{44} p_{32}+p_{33}+p_{24} p_{42}-p_{24} p_{33} p_{42}+p_{23} p_{34} p_{42}+p_{34} p_{43}-p_{33} p_{44}+p_{44}-p_{22} \left(p_{34} p_{43}-p_{33} \left(p_{44}-1\right)+p_{44}-1\right)-1\big]. $$ This expression depends on the $p_{ij}$'s in a very complicated way, which makes simple comparisons such as the one in the OP unlikely to hold.

Let $P=(p_{ij})$ and $Q=(q_{ij})$ be the $n\times n$ transition matrices for the two respective Markov chains. Take $n=5$, $$P=\frac1{1000}\left( \begin{array}{ccccc} 1000 & 0 & 0 & 0 & 0 \\ 1 & 241 & 260 & 38 & 460 \\ 22 & 75 & 283 & 448 & 172 \\ 389 & 67 & 103 & 158 & 283 \\ 0 & 0 & 0 & 0 & 1000 \\ \end{array} \right), $$ $$Q=\frac1{1000}\left( \begin{array}{ccccc} 1000 & 0 & 0 & 0 & 0 \\ 4 & 241 & 259 & 37 & 459 \\ 23 & 531 & 283 & 1 & 162 \\ 390 & 197 & 194 & 158 & 61 \\ 0 & 0 & 0 & 0 & 1000 \\ \end{array} \right). $$

Then $f_{P;21}=\dfrac{30684666}{198426719}=0.154\ldots\not\le0.054\ldots= \dfrac{4510572}{83295877}=f_{Q;21}$, where $f_{P;ij}$ is the probability that the first chain ever reaches a state $j$ from $i$, and $f_{Q;ij}$ is defined similarly.

This disproves your conjecture, because there is no state to the left of $1$.

Let $P=(p_{ij})$ and $Q=(q_{ij})$ be the $n\times n$ transition matrices for the two respective Markov chains. Take $n=5$, $$P=\frac1{1000}\left( \begin{array}{ccccc} 1000 & 0 & 0 & 0 & 0 \\ 1 & 241 & 260 & 38 & 460 \\ 22 & 75 & 283 & 448 & 172 \\ 389 & 67 & 103 & 158 & 283 \\ 0 & 0 & 0 & 0 & 1000 \\ \end{array} \right), $$ $$Q=\frac1{1000}\left( \begin{array}{ccccc} 1000 & 0 & 0 & 0 & 0 \\ 4 & 241 & 259 & 37 & 459 \\ 23 & 531 & 283 & 1 & 162 \\ 390 & 197 & 194 & 158 & 61 \\ 0 & 0 & 0 & 0 & 1000 \\ \end{array} \right). $$

Then $f_{P;21}=\dfrac{30684666}{198426719}=0.154\ldots\not\le0.054\ldots= \dfrac{4510572}{83295877}=f_{Q;21}$, where $f_{P;ij}$ is the probability that the first chain ever reaches a state $j$ from $i$, and $f_{Q;ij}$ is defined similarly.

This disproves your conjecture, because there is no state to the left of $1$.

Remark. Using the formula $f=\lim_{t\uparrow1}(I-tR)^{-1}g$, we see that, for $n=5$, $$f_{P;21}=\big[-p_{24} \left(\left(1-p_{33}\right) p_{41}+p_{31} p_{43}\right)-p_{23} \left(p_{34} p_{41}+p_{31} \left(1-p_{44}\right)\right)+p_{21} \left(-p_{44} p_{33}+p_{33}+p_{34} p_{43}+p_{44}-1\right)\big] \\ \big/\big[p_{23} p_{32}+p_{24} p_{43} p_{32}-p_{23} p_{44} p_{32}+p_{33}+p_{24} p_{42}-p_{24} p_{33} p_{42}+p_{23} p_{34} p_{42}+p_{34} p_{43}-p_{33} p_{44}+p_{44}-p_{22} \left(p_{34} p_{43}-p_{33} \left(p_{44}-1\right)+p_{44}-1\right)-1\big]. $$ This expression depends on the $p_{ij}$'s in a very complicated way, which makes simple comparisons such as the one in the OP unlikely to hold.

Let $P=(p_{ij})$ and $Q=(q_{ij})$ be the $n\times n$ transition matrices for the two respective Markov chains.

The matter becomes much more transparent if we ignore, at least for a moment, the strictness of the inequality restrictions on the $p_{ij}$'s and $q_{ij}$'s. Then we may assume that $n=5$ and $p_{34}=q_{35}=p_{42}=q_{42}=1$. Then clearly $f_{P;3,\le2}=1\not\le0=f_{Q;3,\le2}$, where $f_{P;i,\le j}$ is the probability that the first chain ever reaches a state $\le j$ from $i$, and $f_{Q;i,\le j}$ is defined similarly.

If you still insist on the strictness of the inequality restrictions, then you can have it by the continuity of $f_{P;3,\le2}$ on the appropriate domain.

Let $P=(p_{ij})$ and $Q=(q_{ij})$ be the $n\times n$ transition matrices for the two respective Markov chains. Take $n=5$, $$P=\frac1{1000}\left( \begin{array}{ccccc} 1000 & 0 & 0 & 0 & 0 \\ 1 & 241 & 260 & 38 & 460 \\ 22 & 75 & 283 & 448 & 172 \\ 389 & 67 & 103 & 158 & 283 \\ 0 & 0 & 0 & 0 & 1000 \\ \end{array} \right), $$ $$Q=\frac1{1000}\left( \begin{array}{ccccc} 1000 & 0 & 0 & 0 & 0 \\ 4 & 241 & 259 & 37 & 459 \\ 23 & 531 & 283 & 1 & 162 \\ 390 & 197 & 194 & 158 & 61 \\ 0 & 0 & 0 & 0 & 1000 \\ \end{array} \right). $$

Then $f_{P;21}=\dfrac{30684666}{198426719}=0.154\ldots\not\le0.054\ldots= \dfrac{4510572}{83295877}=f_{Q;21}$, where $f_{P;ij}$ is the probability that the first chain ever reaches a state $j$ from $i$, and $f_{Q;ij}$ is defined similarly.

This disproves your conjecture, because there is no state to the left of $1$.