You misunderstand. P is the co-location event. Events Q and R are both space-like separated from event P. But events Q and R are not space-like separated from each other.
So there exists coordinate frames where P and Q are simultaneous (for example, frame Σ). And there exist coordinate frames where P and R are simultaneous (for example, frame Σ').
Simple: Line j is defined by the equation $$t = t_P$$. Line k is defined by the equation $$t' = t'_P$$. Lines j and k meet at only one event. Calculate the (x,t) or (x', t') of that event.* See that that P is the same P defined as when C' is co-located with M. Thus there is only one place and time where both $$t = t_P$$ and $$t' = t'_P$$ are true.
Events Q and R don't happen in the right place for that to be true. Q.E.D.
Relativity of Simultaneity is demonstrated.
* Assuming chinglu is too lazy to calculate, because we have seen no new calculations since the OP:
- Start with $$t_P = \gamma^{-1} \frac{d'}{c}, \quad t'_P = \frac{d'}{c}, \quad t = t_P = \gamma^{-1} \frac{d'}{c}, \quad t' = t'_P = \frac{d'}{c}, \quad t' = \gamma (t - \beta c^{-1} x)$$ The only way the last three equations can be true is if $$ \frac{d'}{c} = \gamma( \gamma^{-1} \frac{d'}{c} - \beta c^{-1} x) $$ or $$x = 0$$. So the event where $$ t = t_P $$ and $$ t' = t'_P $$ are both true is $$ \left( x = 0, \; t = \gamma^{-1} \frac{d'}{c} \right)$$ which are the frame Σ coordinates of event P as calculated in the OP.
- Alternately, start with $$t = \gamma^{-1} \frac{d'}{c}, \quad t' = \frac{d'}{c}, \quad t = \gamma (t' + \beta c^{-1} x')$$ and solve
$$\gamma^{-1} \frac{d'}{c} = \gamma ( \frac{d'}{c} + \beta c^{-1} x') $$ with solution $$x' = - \beta d'$$. So the event where $$ t = t_P $$ and $$ t' = t'_P $$ are both true is $$ \left( x' = - \beta d', \; t' = \frac{d'}{c} \right)$$ which are the frame Σ' coordinates of event P as calculated in the OP.
