Assume that $char(k) \neq 2,3$ Let $E/K$ be given by a minimal Weierstrass equation of the form $$E: y^2=x^3+Ax +B$$ Prove that $E$ has multiplicative reduction iff $4A^3+27B^2 \in \mathcal{M}$ and $AB \in R^*$.
By Proposition 5.1 in chapter VII Silverman's AEC book we see that $E$ has multiplicative reduction if and only if $v(\Delta)>0$ and $v(c_4)=0$, i.e., $\Delta \in \mathcal{M} \Leftrightarrow 4A^3+27B^2 \in \mathcal{M}$ and $c_4 \in R^* $.
How can I conclude that it is equivalent to $AB \in R^*$?
Well, $\Delta = -16(4A^3 + 27B^2)$ so if $4A^3 + 27B^2$ is in $\mathfrak m$ then $v(\Delta) \geq 1 > 0 $. We also know that $c_4 = -A/27$, so if $AB\in R^*$ then $A\in R^*$ hence $c_4 \in R^*$ and we are done by the proposition.
For the converse, we obviously have $\Delta \in \mathfrak m$ and $A \in R^*$ (examine the work above). But then we know $$-16*27B^2 = \Delta + 16*4A^3$$ and so the LHS has valuation 0 (because the RHS does), hence $B$ has valuation 0 because $v(\Delta) > 0$ and $v(16*4A^3) = 0$.