 Solution

For a nonprincipal Dirichlet character, complete periods sum to zero, so its partial sums are bounded by $q$. Partial summation at $X=q(1+t)$ yieldsThe head is bounded by $1+\log X$, hence $|L(1-it,\overline{\chi })|\ll \log (q+t)$. The completed functional equation givesThe stated gamma bounds make the ratio $O(t^{1/2})$: their exponential factors cancel, and their powers differ by $1/2$. Apply them directly for $t\ge 4$; the compact interval $2\le t\le 4$ is absorbed into the constant. ThereforeFor the conductor-one principal case, Euler summation for zeta at $1-it$, truncated at $X=t^2$, gives a harmonic-size head, a pole term of size $1/t$, and remainder $O((1+t)/X)$. It gives the same $O(\log t)$ bound before applying the zeta functional equation.