In the case of a put option we have $$\partial_x u^{put}_{BS}(\tau,x) =  -K e^{r\tau +x} N(-d_1(\tau,x))
= -K e^{r\tau +x} (1-N(d_1(\tau,x))) = v(\tau, x).$$ Hence the proof of the statement of lemma for the case of a put option is essentially the same as the following argument for a call option.

Using integration by parts and substitution $\xi=d_1(\tau,x)$, we obtain $$\begin{aligned}
\Vert v(\tau,\cdot)\Vert_{L^p}^p 
&=& K^p e^{p r \tau} \int_{-\infty}^\infty e^{p x} (1-N(d_1))^p {\mathrm{d}}x
\\
&\le& 
K^p e^{p r \tau} \int_{-\infty}^\infty e^{p x} (1-N(d_1)) {\mathrm{d}}x
= K^p e^{p r \tau} \int_{-\infty}^\infty \frac{e^{p x}}{p} \frac{e^{-d_1^2/2}}{\sqrt{2\pi}} \frac{1}{\sigma\sqrt{\tau}} {\mathrm{d}}x 
\\
&=& 
K^p e^{p r \tau} \int_{-\infty}^\infty \frac{e^{p \sigma\sqrt{\tau}\xi - p(r+\sigma^2/2)\tau}}{p} \frac{e^{-\xi^2/2}}{\sqrt{2\pi}} {\mathrm{d}}\xi 
= \frac{1}{p} K^p e^{p(p-1)\tau \sigma^2/2}.\end{aligned}$$ Thus $\Vert v(\tau,\cdot)\Vert_{L^p} \le  p^{-1/p} K e^{(p-1) T \sigma^2/2}\equiv C_0$ for any $0<\tau\le T$.

As $\partial_x v = v + w$ where $$w = K e^{ r \tau+x} N'(d_1) \frac{1}{\sigma\sqrt{\tau}}
= K e^{ r \tau+x} \frac{e^{-d_1^2/2}}{\sigma\sqrt{2\pi\tau}}.$$ we obtain $$\begin{aligned}
\Vert w(\tau,\cdot)d_1(\tau,\cdot)^k\Vert_{L^p}^p 
&=& \frac{K^p e^{p r \tau}}{(\sigma\sqrt{2\pi\tau})^{p-1}} \int_{-\infty}^\infty e^{p x} \frac{e^{-p d_1^2/2} |d_1|^{p k}} {\sigma\sqrt{2\pi\tau}} {\mathrm{d}}x \nonumber
\\
&=& 
\frac{K^p e^{p r \tau}}{(\sigma\sqrt{2\pi\tau})^{p-1}} \int_{-\infty}^\infty
e^{p \sigma\sqrt{\tau}\xi - p(r+\sigma^2/2)\tau} \frac{e^{-\xi^2/2} |\xi|^{p k}}{\sqrt{2\pi}} {\mathrm{d}}\xi 
\label{wdineq}
\\
&\le& C_0^p \tau^{-\frac{p-1}{2}} \nonumber\end{aligned}$$ for $k=0,1,2$. Applying (\[wdineq\]) with $k=0$ we obtain $\Vert w(\tau,\cdot)\Vert_{L^p} 