This statement is in BondyMurty text book. every planar graph is 4-vertex-colourable then every plane graph is 4-face-colourable the book said it's a direct consequence of the fact that the dual of a plane graph is planar
so we have that number of edges in a Graph and its dual is rhe same = 2q and the degree of every vertex in dual is number of edges of that field using these facts can we prove this statement?
I'm not sure what you mean by the number of edges being $2q$, but as Bondy & Murty hint the proof only uses the definition of the dual graph (https://en.wikipedia.org/wiki/Dual_graph).
Take a planar graph $P = (V,E)$, with faces $F$ (for some planar embedding). The dual $D$ of $P = (F,I)$ has vertex set $F$ and an edge $i = (f_1,f_2) \in I$ between two vertices $f_1,f_2$ of $F$ whenever they share an edge in $P$ (as faces). Since $P$ is planar $D$ is planar (as Bondy & Murty state), so it must be $4$-vertex-colourable.
Take any proper $4$-vertex-colouring of $D$, that is a colouring of $F$ (as vertices of $D$) such that no adjacent $f_1,f_2$ in $D$ have the same colour. Then, $f_1,f_2$ being adjacent as vertices of $D$ means by definition that faces $f_1,f_2$ share an edge in $P$ as faces. Combining those, this means that we have coloured $F$ such that no two faces of $F$ sharing an edge (in $P$) are coloured the same, which is a proper $4$-face colouring of $P$.