I understood that $T$ is one of the spanning trees which has disjoint edges compared to other $k-1$ edges. then, $T$ has $|V(G)|-1$ edges.
What does the author say in the underlined statement? Isn't it a trivial thing?we know that $|\mathscr P-1|\leq|V(G)|-1$. Is it follow from this inequality and pigeonhole principle? How do I prove the converse?Where will I get the converse?
I draw the example for the illustration of the theorem,
dark blue edged and narrow red edged trees are two edge disjoint trees, what does the theorem say?
The highlighted sentence would be better if it said: whenever $V_1, V_2, \dots, V_p$ is a partition of $G$, and $T$ is a spanning tree of $G$, $T$ must contain at least $p-1$ edges joining distinct parts.
One way to prove this claim is: if we delete all edges of $T$ that join distinct parts among the $V_1, V_2, \dots, V_p$, then $T$ becomes a forest with $p$ components (one on each $V_i$). But deleting one edge can only increase the number of components of a graph by at most $1$, so at least $p-1$ edges must be deleted to get from $1$ to $p$ components.
Another way, as suggested in the comments: consider the graph $T^*$ on $p$ vertices $\{v_1, v_2, \dots, v_p\}$ with an edge $v_iv_j$ whenever $T$ has an edge between $V_i$ and $V_j$. $T^*$ must be connected: if there were no way to get from $\{v_1, \dots, v_k\}$ to $\{v_{k+1}, \dots, v_p\}$ in $T^*$, there would be no way to get from $V_1 \cup \dots \cup V_k$ to $V_{k+1} \cup \dots\cup V_p$ in $T$. So $T^*$ has at least $p-1$ edges.
Once the claim is done, take the $k$ edge-disjoint spanning trees $T_1, T_2, \dots, T_k$ and apply the claim to them. Each $T_i$ has $p-1$ edges joining distinct parts of the partition $V_1, V_2, \dots, V_p$, and these must be different edges for each $T_i$ (because they're edge-disjoint). Altogether, we get $k(p-1)$ edges joining distinct parts, which was what we wanted.