Yes, $p$ is continuous. In proposition 3.7 of this paper it is shown that the lifting map $Map((I,0),(B,b))\to Map((I,0),(E,e))$, $f\mapsto \tilde{f}$ is continuous (and therefore a homeomorphism) when you fix $p(e)=b$. Essentially the same proof will tell you that your map $p$ is continuous, you just don't fix the initial point. Recall the compact open topology is generated by subbasic sets $\langle K,U\rangle=\{f|f(K)\subseteq U\}$ where $K$ is compact and $U$ is open. The key in the proof is to use basic open neighborhoods of the form $$\bigcap_{j=1}^{n}\Big\langle \left[\frac{j-1}{n},\frac{j}{n}\right],U_j\Big\rangle\cap \bigcap_{j=1}^{n-1}\Big\langle \left\{\frac{j}{n}\right\},U_j\cap U_{j+1}\Big\rangle$$ where $n\geq 3$ and $U_j$ evenly covers $\pi(U_j)$.

By using the exponential law, your generalization also has a "yes" answer when $Y=[0,1]^n$ but I think it is not so clear if it holds more generally.

Yes, $p$ is continuous. In proposition 3.7 of this paper it is shown that the lifting map $Map((I,0),(B,b))\to Map((I,0),(E,e))$, $f\mapsto \tilde{f}$ is continuous (and therefore a homeomorphism) when you fix $p(e)=b$. Essentially the same proof will tell you that your map $p$ is continuous, you just don't fix the initial point. Recall the compact open topology is generated by subbasic sets $\langle K,U\rangle=\{f|f(K)\subseteq U\}$ where $K$ is compact and $U$ is open. The key in the proof is to use basic open neighborhoods of the form $$\bigcap_{j=1}^{n}\Big\langle \left[\frac{j-1}{n},\frac{j}{n}\right],U_j\Big\rangle\cap \bigcap_{j=1}^{n-1}\Big\langle \left\{\frac{j}{n}\right\},U_j\cap U_{j+1}\Big\rangle$$ where $n\geq 3$ and $U_j$ evenly covers $\pi(U_j)$.

By using the exponential law, your generalization also has a "yes" answer when $Y=[0,1]^n$ but I think it is not so clear if it holds more generally, for instance, if $Y$ is not a compact metric space.