Part of the excercise I am currently doing is finding the congruences of the following lattice:
The problem I struggle with the most is what happends when $1 \sim d$ - how to find what is the relationship between the rest of the elements?
As there are two votes to close this question as this is somehow marked as offtopic I would like to say that this question is about understanding the concept of congruences in a lattice which as far as I am concerned is strictly connected to this site and complies to the set of questions that can appear on this site. There were asked similar questions on this site before but they are all more complex so I couldn't use them for reference.
The trivial congruences are $\Delta$, the equivalence relation that only says $x\sim x$ for $x\in\{0,a,b,c,d,1\}$, and $\nabla$, the full relation.
Now let $\theta$ be a congruence, and suppose that $0$ is equivalent to some $x\in\{a,b,c\}$ in the lattice. Then since $0\vee y = y$ for all $y$, and $x\vee y = 1$ for all $y\in\{a,b,c\}$ with $x\neq y$, we get that $(1,y)\in\theta$ for all $y\in\{a,b,c\}$ with $y\neq x$. Finally pick two distinct $y,z\in\{a,b,c\}$ different than $x$. Then we have $1\wedge y = y$ and $y\wedge z = 0$, so $(0,y)\in\theta$. So we have that $0,a,b,c,1$ are all $\theta$-equivalent. I let you check that we also have that $d$ is equivalent to $1$, so that $\theta=\nabla$. Dually, one concludes that if $\theta$ contains a pair $(1,x)$ for $x\in\{a,b,c,d\}$, then $\theta=\nabla$ too.
Suppose then that $0$ is not $\theta$-equivalent to any other element of the lattice, and similarly for $1$. You can see that this implies that none of $a,b,c$ can be in the same equivalence class. The only remaining case is that $(c,d)$ is in $\Theta$, all the other blocks being singletons. You can check that this is indeed a congruence by simply enumerating cases.
This finally gives that the congruence lattice of this lattice is a 3-element chain: $$\Delta\subset \theta\subset\nabla$$ where $\theta$ is the equivalence relation given by the blocks $\{c,d\},\{a\},\{b\},\{0\},\{1\}$.
Edit: Since you explicitly asked for the case where $\theta$ contains $(1,d)$ in your question, let me treat it here. Applying $\wedge$ with $(b,b)$, we obtain $(b,0)\in\theta$, and similarly $(a,0)\in\theta$. Since $\theta$ is transitive we have $(b,a)\in\theta$. Applying $\vee$ with $(a,0)$ we obtain $(1,a)\in\theta$. So all of $1,a,b,d,0$ are equivalent. Finally applying $\wedge$ between $(c,c)\in\theta$ and $(d,0)\in\theta$ we obtain $(c,0)\in\theta$. So $\theta$ is the full congruence.