I'm new to using Geometric Brownian Motion, so I'm not sure if what I've done is correct. Be the Geometric Brownian Motion $dS_t = \mu S_tdt + \sigma S_t dW_t$, $H$ a Heaviside, and $p_r, r_k$ constants, then
$$ \begin{align} {\rm I\kern-.3em E} H\Big(\frac{p_r}{S_m}-r_k\Big)\frac{S_T}{S_m}=& {\rm I\kern-.3em E} H\Big(\frac{p_r}{S_m}-r_k\Big){\rm I\kern-.3em E}\frac{S_T}{S_m}\\ =& {\rm I\kern-.3em E} H\Big(\frac{p_r}{S_m}-r_k\Big)e^{(\mu-\sigma^2/2)(T-m)+\sigma^2/2(T+m)}\\ =& {\rm I\kern-.3em E} H\Big(\frac{p_r}{S_m}-r_k\Big)e^{\mu T-m(\mu - \sigma^2)}\\ =& e^{\mu T-m(\mu - \sigma^2)}\int dx \frac{1}{\sigma\sqrt{2\pi m}}e^{\frac{(x-\mu m)^2}{2m\sigma^2}}H\Big(p_rS_0^{-1}e^{-x}-r_k\Big)\\ =& e^{\mu T-m(\mu - \sigma^2)}\int dx \frac{1}{\sigma\sqrt{2\pi m}}e^{\frac{(x-\mu m)^2}{2m\sigma^2}}H\Big(\frac{p_r}{S_0r_k}-e^{x}\Big)\\ =& e^{\mu T-m(\mu - \sigma^2)}\int dx \frac{1}{\sigma\sqrt{2\pi m}}e^{\frac{(x-\mu m)^2}{2m\sigma^2}}H\Big(\log\frac{p_r}{S_0r_k}-x\Big)\\ =& e^{\mu T-m(\mu - \sigma^2)}\int_{-\infty}^{\log\frac{p_r}{S_0r_k}} dx \frac{1}{\sigma\sqrt{2\pi m}}e^{\frac{(x-\mu m)^2}{2m\sigma^2}}\\ \end{align} $$
Change of variables $w=x-\mu m/\sigma\sqrt{m}$ then $dw = dx/\sigma\sqrt{m}$ and $w_+=(\log\frac{p_r}{S_0r_k}-\mu m)/\sigma\sqrt{m}$
\begin{align} {\rm I\kern-.3em E} H\Big(\frac{p_r}{S_m}-r_k\Big)\frac{S_T}{S_m} =& e^{\mu T-m(\mu - \sigma^2)}\int_{-\infty}^{(\log\frac{p_r}{S_0r_k}-\mu m)/\sigma\sqrt{m}} dw \frac{1}{\sqrt{2\pi }}e^{\frac{w^2}{2}}\\ =& e^{\mu T-m(\mu - \sigma^2)}\Phi\Big(\frac{\log\frac{p_r}{S_0r_k}-\mu m}{\sigma\sqrt{m}}\Big)\\ \end{align}
Does it make sense?
I guess you are writing $H(x)=\mathbf{1}_{[0,\infty)}(x)$. We have $$\begin{aligned}H(p_r/S_m-r_k)&=\mathbf{1}_{[0,\infty)}(p_r/S_m-r_k)\\ &=\mathbf{1}_{\{S_m\leq p_r/r_k\}} \end{aligned}$$ We get $$\begin{aligned}E[\mathbf{1}_{\{S_m\leq p_r/r_k\}}(S_T/S_m)|S_m]&=\mathbf{1}_{\{S_m\leq p_r/r_k\}}E[e^{(\mu-\sigma^2/2)(T-m)+\sigma(W_T-W_m)}|S_m]\\ &=\mathbf{1}_{\{S_m\leq p_r/r_k\}}e^{(\mu-\sigma^2/2)(T-m)}e^{\sigma^2(T-m)/2}\\ &=\mathbf{1}_{\{S_m\leq p_r/r_k\}}e^{\mu(T-m)} \end{aligned}$$ And $$\begin{aligned} P(S_m\leq p_r/r_k)&=P\bigg(W_m\leq \frac{\ln(p_r/(r_kS_0))-(\mu-\sigma^2/2)m}{\sigma}\bigg)\\ &=\Phi\bigg(\frac{\ln(p_r/(r_kS_0))-(\mu-\sigma^2/2)m}{\sigma\sqrt{m}}\bigg) \end{aligned}$$ So we conclude: $$E[\mathbf{1}_{\{S_m\leq p_r/r_k\}}(S_T/S_m)]=e^{\mu(T-m)}\Phi\bigg(\frac{\ln(p_r/(r_kS_0))-(\mu-\sigma^2/2)m}{\sigma\sqrt{m}}\bigg)$$