As Ivan Neretin pointed out in the comments, $\pi$ bonds will remain same whether or not in resonance. However, if you have trouble counting the double bond equvalent, you can directly use the formula for it given as $$\ce{DU=\dfrac{2C+2-H+N-X}{2}}$$ so here number of carbons are C=23, H=21, N=1 which also gives 14 as the answer.

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