If a problem Q is NP complete, is another problem reducible to Q also NP complete?
Not necessarily. Any language in $\mathbf{P}$ is poly-time many-one reducible to any non-trivial language (i.e., all except $\varnothing$ or $\Sigma^\ast$); see here, for example. The latter language (i.e., the one you are reducing to) is, in this case, the $\mathbf{NP}$-complete $\mathcal{Q}$. For all we know, your language $\mathcal{P}$ could be any language in $\mathbf{P}$ and, therefore, not necessarily $\mathbf{NP}$-complete (even under the assumption $\mathbf{P} = \mathbf{NP}$).