Assuming the choice of either quarterback [tex](Q)[/tex] or pitcher [tex](P)[/tex] is independent of one another, you have
[tex]\mathbb P(Q\cap P)=\mathbb P(Q)\mathbb P(P)[/tex]
[tex]\dfrac1{315}=\dfrac{\mathbb P(P)}{15}[/tex]
[tex]\implies\mathbb P(P)=\dfrac{15}{315}=\dfrac1{21}[/tex]