The exactly-one-symbol constraints have polynomial size

Introduce a variable $X_{r,c,a}$ for each row $r$, column $c$, and symbol $a\in {\mathrm{\Delta }}_N$, intended to mean that cell $(r,c)$ carries $a$. By [L1], the number of choices for $a$ is a fixed constant, so the total number of such variables is $O(T^2)$.

For each cell, add one at-least-one clause ${\bigvee }_{a\in {\mathrm{\Delta }}_N}{X}_{r,c,a}$ and one pairwise-exclusion clause $\mathrm{\neg }{X}_{r,c,a}\vee \mathrm{\neg }{X}_{r,c,b}$ for each distinct pair $a\mathrm{\ne }b$. Because $\mathrm{\mid }{\mathrm{\Delta }}_N\mathrm{\mid }$ is constant by [L1], each cell contributes only constantly many literals and clauses.

There are $(T+1{)}^2$ cells, and step 2.1 attaches only constant-size data to each one. Therefore the full exactly-one-symbol family has size polynomial in $T$, hence polynomial in the input length once $T$ is polynomially bounded.