A rhombus i

Let the side length of rhombus $ABCD$ be 1; then $DM = \frac 12$ and $NA = \frac 13$ . Let $\angle ABN = \alpha$ and $\angle DCM = \beta$ . Draw a line through $P$ parallel to $AB$ and $CD$ . We note that $x =\alpha + \beta$ . Let $NQ$ be perpendicular to $AB$ and $MR$ be perpendicular to $CD$ , then we have:

$\begin{cases} \tan \alpha = \dfrac {NQ}{BQ} = \dfrac {NA \sin 60^\circ}{1-NA\cos 60^\circ} = \dfrac {\frac 13 \times \frac {\sqrt 3}2}{1-\frac 13\times \frac 12} = \dfrac {\sqrt 3}5 \\ \tan \beta = \dfrac {MR}{CR} = \dfrac {DM \sin 60^\circ}{1+DM\cos 60^\circ} = \dfrac {\frac 12 \times \frac {\sqrt 3}2}{1+\frac 12\times \frac 12} = \dfrac {\sqrt 3}5 \end{cases} \implies \alpha = \beta$

Therefore, $\tan x = \tan (\alpha + \beta) = \tan (2\alpha) = \dfrac {2\tan \alpha}{1-\tan^2 \alpha} = \dfrac {2\times \frac {\sqrt 3}5}{1-\frac 3{25}} = \dfrac {5\sqrt 3}{11} \approx \boxed{0.787}$ .