$\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}$

Take any probability measures $P_0,P_1$ absolutely continuous with respect (w.r.) to $Q$ and to each other. We shall prove the following:

>**Theorem 1.** For any $t\in[0,1]$,
\begin{align*}	\De&:=(1-t)H(P_0)+tH(P_1)-H(P_t) \\ 
&\ge\frac{(1-t)t}2\,\Big[(1-t)\int\Big(\frac{dP_1}{dP_0}-1\Big)^2\,dP_0+t\int\Big(\frac{dP_0}{dP_1}-1\Big)^2\,dP_1\Big] \\
&\ge\frac{(1-t)t}2\,\|P_1-P_0\|^2, 
\end{align*}
where $\|P_1-P_0\|:=\int|dP_1-dP_0|$ is the total variation norm of $P_1-P_0$, 
\begin{equation}
	H(P):=D(P||Q)=\int \ln\frac{dP}{dQ}\,dP,
\end{equation}
and, for any elements $C_0,C_1$ of a linear space, $C_t:=(1-t)C_0+tC_1$. 
Thus, indeed $D(P||Q)$ is strongly convex in $P$ w.r. to the total variation norm. 

We see that the lower bound does not depend on $Q$. 

*Proof of Theorem 1.* Let $f_j:=\frac{dP_j}{dQ}$ for $j=0,1$, so that $f_t=\frac{dP_t}{dQ}$. By Taylor's theorem with the integral form of the remainder, for $h(x):=x\ln x$ and $j=0,1$ we have
\begin{equation}
	h(f_j)=h(f_t)+h'(f_t)(f_j-f_t)+(f_j-f_t)^2\int_0^1 h''((1-s)f_t+sf_j)(1-s)\,ds,
\end{equation}
whence, in view of Jensen's inequality for the convex function $[0,\infty)\ni x\to1/x$ and the probability measure $\mu(ds):=2(1-s)\,ds$ on $[0,1]$,  
\begin{align*}
	\de&:=(1-t)h(f_0)+th(f_1)-h(f_t) \\ 
	&=\frac{(1-t)t}2\,(f_1-f_0)^2\,
	\int_0^1\Big(\frac1{(1-s)f_t+sf_0}+\frac1{(1-s)f_t+sf_1}\Big)2(1-s)\,ds \\ 
	&\ge\frac{(1-t)t}2\,(f_1-f_0)^2\,\Big(\frac1A+\frac1B\Big),
\end{align*}
where 
\begin{align*}
	A&:=\int_0^1[(1-s)f_t+sf_0]2(1-s)\,ds=f_{2t/3}, \\ 
	B&:=\int_0^1[(1-s)f_t+sf_1]2(1-s)\,ds=f_{(2t+1)/3}. 
\end{align*}
So,
\begin{equation}
	\de\ge\frac{(1-t)t}2\,(f_1-f_0)^2\,u(t),
\end{equation}
where 
\begin{multline}
	u(t):=\frac1{f_{2t/3}}+\frac1{f_{(2t+1)/3}}
	\ge(1-t)u(0)+tu(1) \\ 
	=(1-t)\Big(\frac1{f_0}+\frac1{f_{1/3}}\Big)+t\Big(\frac1{f_{2/3}}+\frac1{f_1}\Big) 
	>\frac{1-t}{f_0}+\frac t{f_1}, 
\end{multline}
since $u(t)$ is convex in $t$. So, 
  \begin{multline}
\De=\int\de\,dQ\ge\frac{(1-t)t}2\,\int(f_1-f_0)^2\,\Big(\frac{1-t}{f_0}+\frac t{f_1}\Big)dQ \\ 
=\frac{(1-t)t}2\,\Big[(1-t)\int\Big(\frac{dP_1}{dP_0}-1\Big)^2\,dP_0+t\int\Big(\frac{dP_0}{dP_1}-1\Big)^2\,dP_1\Big],
\end{multline}
which proves the first inequality in Theorem 1. The second inequality follows because, for $f:=\frac{dP_1}{dP_0}$, we have 
\begin{equation}
\int\Big(\frac{dP_1}{dP_0}-1\Big)^2\,dP_0=\int(f-1)^2\,dP_0\ge\Big(\int|f-1|\,dP_0\Big)^2=\|P_1-P_0\|^2	
\end{equation}
and similarly
\begin{equation}
\int\Big(\frac{dP_0}{dP_1}-1\Big)^2\,dP_1\ge\|P_1-P_0\|^2. 
\end{equation}
Theorem 1 is now completely proved.