A property $P$ of topological spaces is said to "pass to quotients" if whenever $p : X \rightarrow Y$ is a quotient map and $X$ has property $P$ then $Y$ has property $P$. For the following properties determine if it "passes to quotients" or give a counter example.
a. Compactness
Compactness passes to quotients, because $p$ is continuous and the image of a compact space (which is $Y$ in this case as $p$ is surjective) is compact.
b. Simply connectedness
The continuous image of a connected space is connected, so $Y$ is also connected. To show that it is simply connected, we need to prove that $\pi_1(Y)=1$. However, $p_{*}: \pi_1(X) \rightarrow \pi_1(Y)$ is defined by $[\omega] \mapsto [p \circ \omega]$. Since $\omega \simeq \text{const}_{pt}$ $\implies p \circ \omega \simeq p \circ \text{const}_{pt} = \text{const}_{p(pt)}$, we have $\pi_1(Y)=1$. So $Y$ is also simply connected.
c. Path connectedness
For any $a,b \in Y$, we have $a' , b' \in X$ such that $p(a') = a$ and $p(b')=b$. Since $X$ is path connected, there exists $\gamma : [0,1] \rightarrow X$ with $\gamma(0) = a'$ and $\gamma(1) = b'$. Hence, we have $p \circ \gamma : [0,1] \rightarrow Y$ with $p \circ \gamma(0) = p(a') = a$ and $p \circ \gamma(1) = p(b') = b$. Since $p$ and $\gamma$ are continuous, so is their composition.
d. Discreteness
I'm not really sure if I understood this one. Does it mean if the set $X$ has the discrete topology, then so must $Y$? If so, then let $a \in Y$. Since $X$ has the discrete topology, $p^{-1}(a)$ is open in $X$. By the definition of a quotient map, $a$ must be open in $Y$. Hence, $Y$ has the discrete topology.
Are my answers correct?
Your answers a, c, and d are correct. However, b is not.
A space is simply connected if it's fundamental group is trivial. While you have correctly observed that a homotopy between paths passes to the quotient, it may be that certain open paths become closed loops in the quotient, hence generators in the fundamental group of the quotient space.
Example. $X = [0, 1]$, $Y = S^1$, $p: X \to Y$ given by identifying the endpoints: $$ p(x) = e^{2 \pi i x}. $$ Now, $\pi_1(X) = 1$, but $\pi_1(Y) \cong \Bbb{Z}$, generated by the identity map.