$\newcommand{\de}{\delta} \newcommand{\ep}{\epsilon}$
Note that $p\ln p-p$ is decreasing in $p\in[0,1]$, so that $p\ln p-p\le q\ln q-q$ and hence $p\ln p-q\ln q\le p-q=|p-q|$ if $0\le q\le p\le1$.
Next, take any real $c\ge1$. Note that $g(p):=p\ln p+cp$ (with $g(0):=0$) is convex in $p\in[0,1]$. So, assuming $0\le p\le q\le1$ and $q\ge e^{-c}$, we have $g(p)\le g(0)\vee g(q)=0\vee g(q)=g(q)$ and hence $p\ln p-q\ln q\le c(q-p)=c|p-q|$. Thus, \begin{equation} p\ln p-q\ln q\le c|p-q| \end{equation} whenever $0\le p,q\le1$ and $q\ge e^{-c}$.
Also, $-q\ln q$ is increasing in $q\in[0,e^{-1}]$ and hence in $q\in[0,e^{-c}]$, so that $-q\ln q\le ce^{-c}$ for $q\in[0,e^{-c}]$. Also, $p\ln p\le0$ if $0\le p\le1$.
Therefore, the difference between the entropies of $Q=(q_i)_{i=1}^N$ and $P=(p_i)_{i=1}^N$ is \begin{equation} \sum_1^N p_i\ln p_i-\sum_1^N q_i\ln q_i=S_1+S_2+S_3, \end{equation} where \begin{align*} S_1 &:=\sum_{i\colon q_i\ge e^{-c}} (p_i\ln p_i-q_i\ln q_i)\le \sum_{i\colon q_i\ge e^{-c}}c|p_i-q_i| \le c\de \quad\text{if}\ \de\ge\|P-Q\|_1, \\ S_2 &:= \sum_{i\colon q_i< e^{-c}}p_i\ln p_i\le0, \\ S_3 &:= \sum_{i\colon q_i< e^{-c}}(-q_i\ln q_i)\le\sum_{i\colon q_i< e^{-c}}ce^{-c}\le Nce^{-c}. \end{align*} So, taking now $c=\ln\frac N\de$ and assuming $N\ge e\de$, we see that \begin{equation} \sum_1^N p_i\ln p_i-\sum_1^N q_i\ln q_i\le c\de+Nce^{-c} =2\de\ln\frac N\de. \end{equation} Taking here $\de=2\ep$ and noting that $N\ge1$, we conclude that \begin{equation} \sum_1^N p_i\ln p_i-\sum_1^N q_i\ln q_i\le 4\ep\ln\frac{N}{2\ep} \end{equation} if $\ep\le1/(2e)$.