If a grammar is regular then it should satisfy pumping lemma for regular grammar

Pumping lemma is For LANGUAGES, Not for Grammars. So, If you meant "If a language is regular then it should satisfy pumping lemma for regular languages" ... Then It is True.

All the Regular languages satisfy Pumping lemma for Regular languages. But it is One way only i.e. Some Non-regular languages may also satisfy Pumping lemma for Regular languages.

for e.g. :

**1. **$L = \left \{ a^*b^*\, \right \}$ : is Regular and All the Regular languages do satisfy Pumping lemma for Regular languages.

**2. **$L = \left \{ w\,\,\,|\,\,w\in \left \{ a,b \right \}^*, n_a(w) = n_b(w) \right \}$ : is Non-regular and Does not satisfy pumping lemma for regular languages. Because whatever Pumping length "P" is taken, There will always be strings which won't be pumped.

**3. **$L = \left \{ a^mb^nc^n\,\,|\,\,m,n\geq1 \right \}\cup \left \{ b,c \right \}^*$ :

is Non-regular But Yet it satisfies the Pumping lemma for Regular languages. Let's rewrite $L$ as

$L = L_1 \cup L_2$ where $L_1 = \left \{ a^mb^nc^n\,\,|\,\,m,n\geq1 \right \}$ and $L_2 = \left \{ b,c \right \}^*$

The language $L$ can be pumped. Since $L_2$ is itself a Regular language, So, Any word from $L_2$ can be pumped arbitrarily and the result is always in $L_2$. Now, See, Any word from $L_1$ can be pumped to any power by choosing a single "$a$" as the pumped part. Usually the result of pumping is in $L_1$, with the exception of words that only contain a single "a" being pumped to the $0^{th}$ power. But if we pump such a word to the $0^{th}$ power, we erase the only "a" and thus we obtain a word from $L_2$

(That very last observation is the whole reason behind $L_2$ being "added" to $L_1$ -- it is there to complete this technical detail.)