My book says, for any $t$ $e^ {tA} = C diag(e^ {tJ_1} ,··· , e^ {tJ_k} ) C^{ −1}$ . Hence,$ |e^ {tA} | ≤ |diag(e^ {tJ_1} ,··· , e^ {tJ_k} ) |$
Where $J_i$ are exponential of jordan blocks of A.
I didn't understand why this has to be true.please help
|A| here denote induced p norm of A
It is not true in general: all you can say is $$ |e^{tA}| \le |C| |C^{-1}| |\text{diag}(e^{t J_1},\ldots,e^{t J_k})|$$ but there's no reason to think $|C| |C^{-1}| = 1$.
You might try some $2 \times 2$ (non-normal) examples. The computations can get rather messy, but you should find that the inequality is false most of the time.
One fairly simple example is $$ A = \pmatrix{1 & 2\cr -1 & -1\cr}$$ This has eigenvalues $\pm i$, and $|e^{tJ}| = 1$ for all $t$. But $$ \exp(tA) = \pmatrix{\cos(t)+\sin(t) & 2 \sin(t)\cr -\sin(t) & \cos(t) - \sin(t) \cr}$$ and since $2 \sin(t)$ can be as much as $2$, it is evident that $|\exp(tA)|$ can be at least $2$. Indeed, for $t=\pi/2$ the norm turns out to be $\sqrt{7/2 + 3 \sqrt{5}/2} \approx 2.618$.