= Solution
First check the <ordered right cosets of a convex lattice subgroup>. Replacing $g,h$ by $p_1g,p_2h$ changes a witness $g\le ph$ into $p_1g\le p_1pp_2^{-1}(p_2h)$, so the relation is independent of representatives. Reflexivity uses $p=1$. If $g\le ph$ and $h\le qk$, two-sided invariance of the <partial order> gives $g\le pqk$, proving transitivity. If both inequalities hold in opposite directions, then
$$
g\le ph\le pqg,\qquad 1\le phg^{-1}\le pq.
$$
Both endpoints are in $P$, so convexity gives $phg^{-1}\in P$ and hence $hg^{-1}\in P$. Thus $Pg=Ph$, proving antisymmetry. The <join> and <meet> formulas are $Pg\vee Ph=P(g\vee h)$ and $Pg\wedge Ph=P(g\wedge h)$. For example, if $g\le pk$ and $h\le qk$, then $g\vee h\le(p\vee q)k$, giving the least-upper-bound property.
The coset order is total exactly when $P$ is a <prime convex lattice subgroup>. If it is total and $a,b\ge1$ have $a\wedge b=1$, assume $Pa\le Pb$. The coset <meet> is then $Pa=P(a\wedge b)=P$, so $a\in P$. Conversely, for $u=gh^{-1}$, its <positive and negative parts in a lattice-ordered group> satisfy $u^+\wedge u^-=1$. Primality puts one of them in $P$. If $u^+\in P$, then $u\le u^+$ gives $Pg\le Ph$; if $u^-\in P$, use $u^{-1}\le u^-$ to obtain $Ph\le Pg$.
It remains to identify the stated intersection condition. If $P$ is prime and $C,D$ properly contain it, choose positive $c\in C\setminus P$ and $d\in D\setminus P$. The cosets $Pc,Pd$ are strictly above $P$. Their <meet> is therefore strictly above $P$, so $c\wedge d\notin P$. But $c\wedge d$ belongs to $C\cap D$ by convexity, showing $C\cap D\ne P$.
If $P$ is not prime, choose disjoint positive $a,b\notin P$. The <principal convex lattice subgroups> $A=\langle a\rangle_c$ and $B=\langle b\rangle_c$ have $A\cap B=\{1\}$. Indeed, disjoint positive elements commute, since $a\vee b=a(a\wedge b)^{-1}b=ab=ba$, and <Riesz decomposition in a lattice-ordered group> shows $a^r\wedge b^s=1$ for every positive $r,s$. Now the distributivity of the <lattice of convex lattice subgroups> gives
$$
(P\vee A)\cap(P\vee B)=P\vee(A\cap B)=P.
$$
Both joins properly contain $P$, violating the intersection condition. This proves the requested equivalence.
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